{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["2"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai7/lv2/img\/4.jpg' \/><\/center>R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c<br\/>$\\left( 4+\\sqrt{15} \\right)\\left( \\sqrt{10}-\\sqrt{6} \\right)\\sqrt{4-\\sqrt{15}}=$_input_ ","hint":"\u0110\u01b0a bi\u1ec3u th\u1ee9c $4+\\sqrt {15}$ v\u00e0o trong d\u1ea5u c\u0103n","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: \u00c1p d\u1ee5ng: Quy t\u1eafc \u0111\u01b0a th\u1eeba s\u1ed1 v\u00e0o trong d\u1ea5u c\u0103n<br\/>B\u01b0\u1edbc 2: \u0110\u01b0a bi\u1ec3u th\u1ee9c $\\sqrt 10-\\sqrt 6$ v\u00e0o trong d\u1ea5u c\u0103n v\u00e0 \u0111\u01b0a bi\u1ec3u th\u1ee9c trong c\u0103n v\u1ec1 d\u1ea1ng $(A+B)^2$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/>$\\begin{align} & \\,\\,\\,\\left( 4+\\sqrt{15} \\right)\\left( \\sqrt{10}-\\sqrt{6} \\right)\\sqrt{4-\\sqrt{15}} \\\\ & =\\sqrt{{{\\left( 4+\\sqrt{15} \\right)}^{2}}\\left( 4-\\sqrt{15} \\right)}\\left( \\sqrt{10}-\\sqrt{6} \\right) \\\\ & =\\sqrt{\\left( {{4}^{2}}-\\sqrt{{{15}^{2}}} \\right)\\left( 4+\\sqrt{15} \\right){{\\left( \\sqrt{10}-\\sqrt{6} \\right)}^{2}}} \\\\ & =\\sqrt{\\left( 4+\\sqrt{15} \\right){{\\left[ \\sqrt{2}\\left( \\sqrt{5}-\\sqrt{3} \\right) \\right]}^{2}}} \\\\ & =\\sqrt{2\\left( 4+\\sqrt{15} \\right){{\\left( \\sqrt{5}-\\sqrt{3} \\right)}^{2}}} \\\\ & =\\sqrt{\\left( 8+2\\sqrt{15} \\right){{\\left( \\sqrt{5}-\\sqrt{3} \\right)}^{2}}} \\\\ & =\\sqrt{{{\\left( \\sqrt{5}+\\sqrt{3} \\right)}^{2}}{{\\left( \\sqrt{5}-\\sqrt{3} \\right)}^{2}}} \\\\ & =\\sqrt{{{\\left( 5-3 \\right)}^{2}}}\\\\&=2 \\\\ \\end{align}$ <br\/> <span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $2$<\/span><\/span><\/span>"}]}],"id_ques":731},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<span class='basic_left'>V\u1edbi $x\\ge 0;\\,\\,x\\ne 9$. Cho hai bi\u1ec3u th\u1ee9c<br\/>$A=\\dfrac{x+3}{x-9}+\\dfrac{2}{\\sqrt{x}+3}-\\dfrac{1}{\\sqrt{x}-3}$ v\u00e0 $B=\\dfrac{\\sqrt{x}+1}{\\sqrt{x}+2}$ <br\/><b> C\u00e2u 1: <\/b>R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $A$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0:<\/span> ","select":["A. $\\dfrac{\\sqrt{x}-2}{\\sqrt{x}-3}$ ","B. $\\dfrac{\\sqrt{x}-2}{\\sqrt{x}+3}$ ","C. $\\dfrac{\\sqrt{x}+2}{\\sqrt{x}-3}$","D. $\\dfrac{\\sqrt{x}+2}{\\sqrt{x}+3}$"],"hint":"Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c c\u00e1c ph\u00e2n th\u1ee9c v\u1edbi m\u1eabu th\u1ee9c chung $(\\sqrt{x}+3)(\\sqrt{x}-3)$","explain":"<span class='basic_left'>V\u1edbi $x\\ge 0;\\,\\,x\\ne 9$. Ta c\u00f3:<br\/> $\\begin{align} A&=\\dfrac{x+3}{x-9}+\\dfrac{2}{\\sqrt{x}+3}-\\dfrac{1}{\\sqrt{x}-3} \\\\ & =\\dfrac{x+3+2\\left( \\sqrt{x}-3 \\right)-\\left( \\sqrt{x}+3 \\right)}{\\left( \\sqrt{x}+3 \\right)\\left( \\sqrt{x}-3 \\right)} \\\\ & =\\dfrac{x+3+2\\sqrt{x}-6-\\sqrt{x}-3}{\\left( \\sqrt{x}+3 \\right)\\left( \\sqrt{x}-3 \\right)} \\\\ & =\\dfrac{x+\\sqrt{x}-6}{\\left( \\sqrt{x}+3 \\right)\\left( \\sqrt{x}-3 \\right)} \\\\ & =\\dfrac{x-2\\sqrt{x}+3\\sqrt x -6}{\\left( \\sqrt{x}+3 \\right)\\left( \\sqrt{x}-3 \\right)} \\\\ & =\\dfrac{\\left( \\sqrt{x}-2 \\right)\\left( \\sqrt{x}+3 \\right)}{\\left( \\sqrt{x}-3 \\right)\\left( \\sqrt{x}+3 \\right)} \\\\ & =\\dfrac{\\sqrt{x}-2}{\\sqrt{x}-3} \\\\ \\end{align}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span>","column":2}]}],"id_ques":732},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"<span class='basic_left'>V\u1edbi $x\\ge 0;\\,\\,x\\ne 9$. Cho hai bi\u1ec3u th\u1ee9c<br\/>$A=\\dfrac{x+3}{x-9}+\\dfrac{2}{\\sqrt{x}+3}-\\dfrac{1}{\\sqrt{x}-3}$ v\u00e0 $B=\\dfrac{\\sqrt{x}+1}{\\sqrt{x}+2}$ <br\/><b> C\u00e2u 2: <\/b> T\u00ecm $x$ \u0111\u1ec3 $A < 1$<\/span> ","select":["A. $x \\ge 0$ ","B. $x > 9$ ","C. $0\\le x < 9$","D. $x < 9$"],"hint":"Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh $A<1$","explain":"<span class='basic_left'>Theo c\u00e2u 1, v\u1edbi $x\\ge 0;\\,\\,x\\ne 9$, ta c\u00f3: <br\/> $A=\\dfrac{\\sqrt{x}-2}{\\sqrt{x}-3}$ <br\/>$\\begin{align} A<1&\\Leftrightarrow \\dfrac{\\sqrt{x}-2}{\\sqrt{x}-3}<1 \\\\ & \\Leftrightarrow \\dfrac{\\sqrt{x}-2-\\sqrt{x}+3}{\\sqrt{x}-3}<0 \\\\ & \\Leftrightarrow \\dfrac{1}{\\sqrt{x}-3}<0 \\\\ & \\Leftrightarrow \\sqrt{x}-3<0 \\,\\,\\, (\\text {V\u00ec} 1>0)\\\\ & \\Leftrightarrow x<9 \\\\ \\end{align}$ <br\/>M\u00e0 $x\\ge 0;\\,\\,x\\ne 9$. Do \u0111\u00f3 A < 1 $\\Leftrightarrow 0\\le x<9$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span>","column":2}]}],"id_ques":733},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{1}{4}$","B. $\\dfrac{3}{4}$","C. $\\dfrac{5}{4}$"],"ques":" <span class='basic_left'>V\u1edbi $x\\ge 0;\\,\\,x\\ne 9$. Cho hai bi\u1ec3u th\u1ee9c<br\/>$A=\\dfrac{x+3}{x-9}+\\dfrac{2}{\\sqrt{x}+3}-\\dfrac{1}{\\sqrt{x}-3}$ v\u00e0 $B=\\dfrac{\\sqrt{x}+1}{\\sqrt{x}+2}$ <br\/>V\u1edbi $A = B$ th\u00ec $x=$?<\/span>","hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $A=B$ v\u1edbi bi\u1ec3u th\u1ee9c $A$ l\u00e0 k\u1ebft qu\u1ea3 r\u00fat g\u1ecdn c\u00e2u 1","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: Quy \u0111\u1ed3ng v\u00e0 kh\u1eed m\u1eabu ph\u01b0\u01a1ng tr\u00ecnh $A=B$<br\/>B\u01b0\u1edbc 2: R\u00fat g\u1ecdn ph\u01b0\u01a1ng tr\u00ecnh v\u00e0 t\u00ecm nghi\u1ec7m<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Theo c\u00e2u 1, v\u1edbi $x\\ge 0;\\,\\,x\\ne 9$, ta c\u00f3: <br\/>$A=\\dfrac{\\sqrt{x}-2}{\\sqrt{x}-3}$ <br\/>$\\begin{align} A=B&\\Leftrightarrow \\dfrac{\\sqrt{x}-2}{\\sqrt{x}-3}=\\dfrac{\\sqrt{x}+1}{\\sqrt{x}+2} \\\\ & \\Leftrightarrow \\dfrac{\\left( \\sqrt{x}-2 \\right)\\left( \\sqrt{x}+2 \\right)}{\\left( \\sqrt{x}-3 \\right)\\left( \\sqrt{x}+2 \\right)}=\\dfrac{\\left( \\sqrt{x}+1 \\right)\\left( \\sqrt{x}-3 \\right)}{\\left( \\sqrt{x}+2 \\right)\\left( \\sqrt{x}-3 \\right)} \\\\ & \\Rightarrow \\left( \\sqrt{x}-2 \\right).\\left( \\sqrt{x}+2 \\right)=\\left( \\sqrt{x}+1 \\right).\\left( \\sqrt{x}-3 \\right) \\\\ & \\Leftrightarrow x-4=x-2\\sqrt{x}-3 \\\\ & \\Leftrightarrow 2\\sqrt{x}=1 \\\\ & \\Leftrightarrow \\sqrt{x}\\,\\,\\,=\\dfrac{1}{2} \\\\ & \\Leftrightarrow x\\,\\,\\,\\,\\,\\,\\,\\,=\\dfrac{1}{4} \\,\\, (\\text {th\u1ecfa m\u00e3n})\\\\ \\end{align}$<\/span><\/span>"}]}],"id_ques":734},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"K\u1ebft qu\u1ea3 r\u00fat g\u1ecdn c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{1}{\\sqrt[3]{9}+\\sqrt[3]{15}+\\sqrt[3]{25}}+\\dfrac{\\sqrt[3]{81}}{3}$ l\u00e0: ","select":["A. $\\sqrt[3]{5}$ ","B. $\\sqrt[3]{5}+\\sqrt[3]{3}$ ","C. $\\dfrac{\\sqrt[3]{5}-\\sqrt[3]{3}}{2}$","D. $\\dfrac{\\sqrt[3]{5}+\\sqrt[3]{3}}{2}$"],"hint":"Tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu v\u00e0 \u00e1p d\u1ee5ng quy t\u1eafc \u0111\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: Nh\u00e2n c\u1ea3 t\u1eed v\u00e0 m\u1eabu c\u1ee7a ph\u00e2n th\u1ee9c th\u1ee9 nh\u1ea5t v\u1edbi bi\u1ec3u th\u1ee9c li\u00ean h\u1ee3p c\u1ee7a m\u1eabu.<br\/>B\u01b0\u1edbc 2: Quy \u0111\u1ed3ng v\u00e0 r\u00fat g\u1ecdn<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<br\/><\/span>Ta c\u00f3:<br\/>$ \\,\\,\\,\\,\\,\\dfrac{1}{\\sqrt[3]{9}+\\sqrt[3]{15}+\\sqrt[3]{25}}+\\dfrac{\\sqrt[3]{81}}{3}$<br\/>$=\\dfrac{\\left( \\sqrt[3]{5}-\\sqrt[3]{3} \\right)}{\\left( \\sqrt[3]{9}+\\sqrt[3]{15}+\\sqrt[3]{25} \\right)\\left( \\sqrt[3]{5}-\\sqrt[3]{3} \\right)}\\,$$+\\dfrac{3\\sqrt[3]{3}}{3} $<br\/>$ =\\dfrac{\\sqrt[3]{5}-\\sqrt[3]{3}}{\\sqrt[3]{{{5}^{3}}}-\\sqrt[3]{{{3}^{3}}}}+\\sqrt[3]{3} $<br\/>$=\\dfrac{\\sqrt[3]{5}-\\sqrt[3]{3}}{2}+\\sqrt[3]{3} $<br\/>$ =\\dfrac{\\sqrt[3]{5}+\\sqrt[3]{3}}{2} $ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><br\/><i>L\u01b0u \u00fd<\/i>$\\dfrac{1}{\\sqrt[3]{{{a}^{2}}}+\\sqrt[3]{ab}+\\sqrt[3]{{{b}^{2}}}}\\,$$=\\dfrac{\\sqrt[3]{a}-\\sqrt[3]{b}}{\\left( \\sqrt[3]{a}-\\sqrt[3]{b} \\right)\\left( \\sqrt[3]{{{a}^{2}}}+\\sqrt[3]{ab}+\\sqrt[3]{{{b}^{2}}} \\right)}\\,$$=\\dfrac{\\sqrt[3]{a}-\\sqrt[3]{b}}{a-b}$ <\/span>","column":2}]}],"id_ques":735},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"select":["A. $\\sqrt{7}$","B. $2+\\sqrt{7}$","C. $4+\\sqrt{7}$"],"ques":"R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c <br\/>$\\sqrt{16-6\\sqrt{7}}+\\sqrt{29+4\\sqrt{7}}=$?","hint":"Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c trong c\u0103n v\u1ec1 d\u1ea1ng $(a \\pm b)^2$ ","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\,\\,\\,\\sqrt{16-6\\sqrt{7}}+\\sqrt{29+4\\sqrt{7}}$<br\/>$ =\\sqrt{{{3}^{2}}-2.3.\\sqrt{7}+\\sqrt{{{7}^{2}}}}\\,$$+\\sqrt{{{\\left( 2\\sqrt{7} \\right)}^{2}}+2.2\\sqrt{7}.1+{{1}^{2}}}$<br\/>$ =\\sqrt{{{\\left( 3-\\sqrt{7} \\right)}^{2}}}+\\sqrt{{{\\left( 2\\sqrt{7}+1 \\right)}^{2}}}$<br\/>$=3-\\sqrt{7}+2\\sqrt{7}+1\\,\\,\\,\\,(\\text {V\u00ec}\\,\\,3\\,>\\sqrt{7}) $<br\/>$ =4+\\sqrt{7}$<br\/><i>Ghi nh\u1edb:<\/i> V\u1edbi $A$ l\u00e0 m\u1ed9t bi\u1ec3u th\u1ee9c, ta c\u00f3 $\\sqrt{A^2}=|A|=\\left\\{ \\begin{align} & A\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,\\,\\,\\,\\,\\,\\,{A}\\ge {0} \\\\ & -A\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,\\,\\,\\,\\,\\,\\,{A}<{0} \\\\\\end{align} \\right.$<\/span> "}]}],"id_ques":736},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai7/lv2/img\/9.jpg' \/><\/center>V\u1edbi $a > 0; b > 0$ v\u00e0 $a\\ne b$. R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c:<br\/>$\\dfrac{a+b-2\\sqrt{ab}}{\\sqrt{a}-\\sqrt{b}}:\\dfrac{1}{\\sqrt{a}+\\sqrt{b}}=a-b$ ","select":["A. \u0110\u00fang","B. Sai"],"hint":"\u0110\u01b0a t\u1eed th\u1ee9c c\u1ee7a ph\u00e2n th\u1ee9c b\u1ecb chia v\u1ec1 d\u1ea1ng $(A-B)^2$ v\u00e0 r\u00fat g\u1ecdn v\u1edbi m\u1eabu th\u1ee9c","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: Bi\u1ebfn \u0111\u1ed5i v\u1ebf tr\u00e1i<br\/>B\u01b0\u1edbc 2: \u00c1p d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $(a-b)(a+b)=a^2-b^2$ v\u00e0 r\u00fat g\u1ecdn.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<br\/><\/span>Ta c\u00f3:<br\/> $\\begin{align}\\text {V\u1ebf tr\u00e1i}=&\\,\\,\\,\\,\\,\\dfrac{a+b-2\\sqrt{ab}}{\\sqrt{a}-\\sqrt{b}}:\\dfrac{1}{\\sqrt{a}+\\sqrt{b}}\\\\&=\\dfrac{{{\\left( \\sqrt{a}-\\sqrt{b} \\right)}^{2}}}{\\sqrt{a}-\\sqrt{b}}.\\left( \\sqrt{a}+\\sqrt{b} \\right)\\\\&=\\left( \\sqrt{a}-\\sqrt{b} \\right)\\left( \\sqrt{a}+\\sqrt{b} \\right)\\\\&=a-b=\\text{V\u1ebf ph\u1ea3i} \\\\ \\end{align}$<br\/> Do \u0111\u00f3 kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0110\u00fang <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><\/span>","column":2}]}],"id_ques":737},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank_random","correct":[[["-3"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai7/lv2/img\/1.png' \/><\/center>Bi\u1ebft $\\sqrt{13-4\\sqrt{3}}=a+b\\sqrt{3}$ (V\u1edbi $a, b \\in \\mathbb Z$)<br\/>Khi \u0111\u00f3 $a-b=$_input_","hint":"Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c d\u01b0\u1edbi d\u1ea5u c\u0103n v\u1ec1 $(a \\pm b)^2$, r\u1ed3i s\u1eeda d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $\\sqrt {A^2}=|A|$ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c d\u01b0\u1edbi d\u1ea5u c\u0103n v\u1ec1 $(2 \\sqrt{3}-1)^{2}$<br\/>B\u01b0\u1edbc 2: \u00c1p d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $\\sqrt {A^2}=|A|$ v\u00e0 \u0111\u1ed3ng nh\u1ea5t h\u1ec7 s\u1ed1 \u0111\u1ec3 t\u00ecm $a, b$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: <br\/>$\\begin{align}&\\,\\,\\,\\,\\,\\sqrt{13-4\\sqrt{3}}\\\\&=\\sqrt{{{\\left( 2\\sqrt{3} \\right)}^{2}}-2.2\\sqrt{3}.1+{{1}^{2}}}\\\\&=\\sqrt{{{\\left( 2\\sqrt{3}-1 \\right)}^{2}}}\\\\&=2\\sqrt{3}-1\\,\\,\\,(\\text {V\u00ec}\\,\\,\\, 2\\sqrt{3}\\, >\\, 1 )\\\\ \\end{align}$<br\/>M\u00e0 $\\sqrt{13-4\\sqrt{3}}=a+b\\sqrt{3}=2\\sqrt{3}-1$<br\/>V\u00ec $a,b \\in \\mathbb Z $ n\u00ean ta c\u00f3: $a= -1;\\, b=2\\,$.<br\/> Do \u0111\u00f3 $ \\,a-b=-1-2=-3$ <br\/> <span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $-3$<\/span><\/span><\/span>"}]}],"id_ques":738},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $A_{min} =-\\dfrac{17}{4}$ khi $a =\\dfrac{25}{4}$","B. $A_{min} =-\\dfrac{15}{4}$ khi $a =\\dfrac{1}{4}$","C. $A_{min} =-\\dfrac{13}{4}$ khi $a =\\dfrac{1}{2}$"],"ques":"Cho $a \\ge0$. T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c $A=a-5\\sqrt{a}+2$<br\/>\u0110\u00e1p s\u1ed1: $A_{\\min}=$?khi $a =$ ?<br\/>(K\u1ebft qu\u1ea3 vi\u1ebft d\u01b0\u1edbi d\u1ea1ng ph\u00e2n s\u1ed1)","hint":"Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c $A$ v\u1ec1 d\u1ea1ng $f(a)^2+b$, t\u1eeb \u0111\u00f3 \u0111\u00e1nh gi\u00e1 $A$ \u0111\u1ec3 t\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t. ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c $A$ v\u1ec1 d\u1ea1ng ${\\left( \\sqrt{a}-\\dfrac{5}{2} \\right)}^{2}-\\dfrac{17}{4}$<br\/>B\u01b0\u1edbc 2: \u0110\u00e1nh gi\u00e1 v\u00e0 t\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $A$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/>$A=a-5\\sqrt{a}+2\\,$<br\/>$=\\sqrt{{{a}^{2}}}-2.\\dfrac{5}{2}.\\sqrt{a}+ {{\\left( \\dfrac{5}{2} \\right)}^{2}}-\\dfrac{25}{4}+2 \\,$<br\/>$={{\\left( \\sqrt{a}-\\dfrac{5}{2}\\right)}^{2}}-\\dfrac{17}{4}$<br\/>V\u00ec ${{\\left( \\sqrt{a}-\\dfrac{5}{2}\\right)}^{2}}\\ge 0$ v\u1edbi m\u1ecdi $a\\ge 0$ <br\/>N\u00ean $A={{\\left( \\sqrt{a}-\\dfrac{5}{2} \\right)}^{2}}-\\dfrac{17}{4}\\ge -\\dfrac{17}{4}$ v\u1edbi m\u1ecdi $a\\ge 0$<br\/>Do \u0111\u00f3 ${{A}_{\\min }}=-\\dfrac{17}{4}\\Leftrightarrow {{\\left( \\sqrt{a}-\\dfrac{5}{2} \\right)}^{2}}=0\\,$$\\Leftrightarrow \\sqrt{a}=\\dfrac{5}{2}\\Leftrightarrow a=\\dfrac{25}{4}$ <\/span> <\/span> <\/span> "}]}],"id_ques":739},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank_random","correct":[[["2"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai7/lv2/img\/9.jpg' \/><\/center>Th\u1ef1c hi\u1ec7n ph\u00e9p t\u00ednh<br\/>$\\dfrac{\\left( \\sqrt{3}+1 \\right)\\left( 4-2\\sqrt{3} \\right)}{\\sqrt{3}-1}$=_input_","hint":"\u0110\u01b0a $4-2 \\sqrt3$ v\u1ec1 d\u1ea1ng $(a-b)^2$, sau \u0111\u00f3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{align} & \\,\\,\\,\\dfrac{\\left( \\sqrt{3}+1 \\right)\\left( 4-2\\sqrt{3} \\right)}{\\sqrt{3}-1} \\\\ & =\\dfrac{\\left( \\sqrt{3}+1 \\right){{\\left( \\sqrt{3}-1 \\right)}^{2}}}{\\sqrt{3}-1} \\\\ & =\\left( \\sqrt{3}+1 \\right)\\left( \\sqrt{3}-1 \\right) \\\\ & =2 \\\\ \\end{align}$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $2$<\/span><\/span> "}]}],"id_ques":740},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank_random","correct":[[["39"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh:$\\dfrac{2}{3}\\sqrt{9x-27}+\\sqrt{x-3}\\,$$=6+\\sqrt{4x-12}$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${_input_}","hint":"\u00c1p d\u1ee5ng: Quy t\u1eafc \u0111\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: \u0110\u1eb7t \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 ngh\u0129a: $\\sqrt{A}$ c\u00f3 ngh\u0129a $\\Leftrightarrow A \\ge 0$<br\/>B\u01b0\u1edbc 2: \u00c1p d\u1ee5ng: Quy t\u1eafc \u0111\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n : $\\sqrt{A^2B}=|A|\\sqrt{B}$ v\u1edbi $B\\ge 0$<br\/>B\u01b0\u1edbc 3: R\u00fat g\u1ecdn c\u00e1c c\u0103n th\u1ee9c \u0111\u1ed3ng d\u1ea1ng <br\/>B\u01b0\u1edbc 4. Bi\u1ebfn \u0111\u1ed5i ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng $\\sqrt{A}=B\\Leftrightarrow A=B^2$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ge 3$<br\/>Ta c\u00f3:<br\/>$\\,\\,\\,\\,\\,\\dfrac{2}{3}\\sqrt{9x-27}+\\sqrt{x-3}=6+\\sqrt{4x-12}$<br\/>$\\Leftrightarrow \\dfrac{2}{3}\\sqrt{9(x-3)}+\\sqrt{x-3}=6+\\sqrt{4(x-3)}$<br\/>$ \\Leftrightarrow 2\\sqrt{x-3}+\\sqrt{x-3}-2\\sqrt{x-3}\\,$$=6$<br\/>$ \\Leftrightarrow \\sqrt{x-3}=6 $<br\/>$ \\Leftrightarrow x-3\\,\\,\\,\\,\\,=36 $<br\/>$ \\Leftrightarrow x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=39\\,\\,\\,\\text{(th\u1ecfa m\u00e3n)} $ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{39\\}$ <br\/> <span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $39$<\/span><\/span>"}]}],"id_ques":741},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<span class='basic_left'>V\u1edbi $x \\ge 0$ v\u00e0 $x \\ne 9$. Cho bi\u1ec3u th\u1ee9c $M=\\left( \\dfrac{2}{\\sqrt{x}-3}+\\dfrac{1}{\\sqrt{x}+3} \\right)\\,$$:\\dfrac{\\sqrt{x}+1}{\\sqrt{x}-3}$ <br\/><b>C\u00e2u 1: <\/b> R\u00fat g\u1ecdn $M$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0:<\/span> ","select":["A. $-\\dfrac{3}{\\sqrt{x}+3}$ ","B. $\\dfrac{3}{\\sqrt{x}+3}$ ","C. $\\dfrac{\\sqrt{x}+1}{\\sqrt{x}+3}$","D. $\\dfrac{\\sqrt{x}+1}{\\sqrt{x}-3}$"],"hint":"R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c trong ngo\u1eb7c tr\u01b0\u1edbc: M\u1eabu th\u1ee9c chung $(\\sqrt{x}-3)(\\sqrt{x}+3)$","explain":"<span class='basic_left'>V\u1edbi $x \\ge 0$ v\u00e0 $x \\ne 9$, ta c\u00f3: <br\/>$\\begin{align} M&=\\left( \\dfrac{2}{\\sqrt{x}-3}+\\dfrac{1}{\\sqrt{x}+3} \\right):\\dfrac{\\sqrt{x}+1}{\\sqrt{x}-3} \\\\ & =\\dfrac{2\\sqrt{x}+6+\\sqrt{x}-3}{\\left( \\sqrt{x}-3 \\right)\\left( \\sqrt{x}+3 \\right)}.\\dfrac{\\sqrt{x}-3}{\\sqrt{x}+1} \\\\ & =\\dfrac{3\\sqrt{x}+3}{\\sqrt{x}+3}.\\dfrac{1}{\\sqrt{x}+1} \\\\ & =\\dfrac{3\\left( \\sqrt{x}+1 \\right)}{\\sqrt{x}+3}.\\dfrac{1}{\\sqrt{x}+1} \\\\ & =\\dfrac{3}{\\sqrt{x}+3} \\\\ \\end{align}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span>","column":2}]}],"id_ques":742},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"select":["A. $5 + 3\\sqrt{3}$","B. $3\\sqrt{3}$","C. $6-3\\sqrt{3}$"],"ques":" <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $M=\\left( \\dfrac{2}{\\sqrt{x}-3}+\\dfrac{1}{\\sqrt{x}+3} \\right):\\dfrac{\\sqrt{x}+1}{\\sqrt{x}-3}\\,\\,\\,$(V\u1edbi $x \\ge 0$ v\u00e0 $x \\ne 9$)<br\/> V\u1edbi $x=4-2\\sqrt{3}$ th\u00ec $M=$?<\/span> ","hint":"Bi\u1ebfn \u0111\u1ed5i gi\u00e1 tr\u1ecb c\u1ee7a $x$ v\u1ec1 d\u1ea1ng $(a-b)^2$ v\u00e0 t\u00ednh $\\sqrt x$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c $4-2\\sqrt{3}=(\\sqrt{3}-1)^2$<br\/>B\u01b0\u1edbc 2: Thay $x$ v\u00e0o t\u00ednh $M$<br\/>B\u01b0\u1edbc 3: Tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Theo c\u00e2u 1, v\u1edbi $x \\ge 0$ v\u00e0 $x \\ne 9$, ta c\u00f3:<br\/> $M=\\dfrac{3}{\\sqrt{x}+3}$ <br\/>Theo \u0111\u1ec1 b\u00e0i ta c\u00f3:<br\/> $x=4-2\\sqrt{3}\\Rightarrow\\sqrt{x}=\\sqrt{4-2\\sqrt{3}}=\\sqrt{{{\\left( \\sqrt{3}-1 \\right)}^{2}}}\\,$$=\\sqrt{3}-1$<br\/>Thay $\\sqrt{x}=\\sqrt{3}-1$ v\u00e0o $M$ ta \u0111\u01b0\u1ee3c: <br\/>$M=\\dfrac{3}{\\sqrt{3}-1+3}=\\dfrac{3}{\\sqrt{3}+2}\\,$$=\\dfrac{3\\left( 2-\\sqrt{3} \\right)}{4-(\\sqrt{3})^2}=6-3\\sqrt{3}$<\/span><\/span> "}]}],"id_ques":743},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"fill_the_blank_random","correct":[[["0"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $M=\\left( \\dfrac{2}{\\sqrt{x}-3}+\\dfrac{1}{\\sqrt{x}+3} \\right):\\dfrac{\\sqrt{x}+1}{\\sqrt{x}-3}\\,\\,\\,$(v\u1edbi $x \\ge 0$ v\u00e0 $x \\ne 9$)<br\/> <b> C\u00e2u 3: <\/b> T\u00ecm $x\\in \\mathbb Z$ \u0111\u1ec3 $M\\in \\mathbb Z$<br\/>\u0110\u00e1p s\u1ed1: $x=$_input_<\/span>","hint":"Cho m\u1eabu c\u1ee7a ph\u00e2n th\u1ee9c \u0111\u00e3 r\u00fat g\u1ecdn b\u1eb1ng \u01b0\u1edbc c\u1ee7a t\u1eed ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: T\u00ecm $x$ \u0111\u1ec3 $\\sqrt{x}+3 \\in \u01af(3)$ <br\/>B\u01b0\u1edbc 2: So s\u00e1nh v\u1edbi \u0111i\u1ec1u ki\u1ec7n v\u00e0 k\u1ebft lu\u1eadn<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'>Theo c\u00e2u 1, v\u1edbi $x \\ge 0$ v\u00e0 $x \\ne 9$, ta c\u00f3:<br\/> $M=\\dfrac{3}{\\sqrt{x}+3}$ <br\/>$M\\in \\mathbb Z\\Leftrightarrow \\dfrac{3}{\\sqrt{x}+3}\\in \\mathbb Z\\Leftrightarrow \\sqrt{x}+3\\,\\,$ thu\u1ed9c $\u01af (3)=\\left\\{ \\pm 1;\\pm 3 \\right\\}$<br\/>Ta c\u00f3 b\u1ea3ng sau: <br\/><table> <tr> <th>$\\sqrt{x}+3$<\/th> <th>$-3$<\/th> <th>$-1$<\/th> <th>$1$<\/th> <th>$3$<\/th> <\/tr> <tr> <td>$\\sqrt{x}$<\/td> <td>$-6$<\/td> <td>$-4$<\/td> <td>$-2$<\/td> <td>$0$<\/td> <\/tr> <tr> <td>$x$<\/td> <td>lo\u1ea1i <\/td> <td>lo\u1ea1i <\/td> <td>lo\u1ea1i <\/td> <td>$0$<\/td> <\/tr><\/table> <br\/>V\u1eady $x= 0$ th\u00ec $M\\in \\mathbb Z$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $0$<\/span><\/span><\/span>"}]}],"id_ques":744},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{7}{2}$","B. $\\dfrac{5}{2}$","C. $\\dfrac{3}{2}$"],"ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh:$\\sqrt{8x-12}+\\sqrt{18x-27}\\,$$=12-\\sqrt{2x-3}$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${?}","hint":"\u00c1p d\u1ee5ng quy t\u1eafc \u0111\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: \u0110\u1eb7t \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 ngh\u0129a: $\\sqrt{A}$ c\u00f3 ngh\u0129a $\\Leftrightarrow A \\ge 0$<br\/>B\u01b0\u1edbc 2: \u00c1p d\u1ee5ng: Quy t\u1eafc \u0111\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n : $\\sqrt{A^2B}=|A|\\sqrt{B}$ v\u1edbi $B\\ge 0$<br\/>B\u01b0\u1edbc 3: R\u00fat g\u1ecdn c\u00e1c c\u0103n th\u1ee9c \u0111\u1ed3ng d\u1ea1ng <br\/>B\u01b0\u1edbc 4. Bi\u1ebfn \u0111\u1ed5i ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng $\\sqrt{A}=B\\Leftrightarrow A=B^2$ v\u1edbi $B\\ge 0$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ge\\dfrac{3}{2}$<br\/>Ta c\u00f3:<br\/>$ \\,\\,\\,\\,\\,\\sqrt{8x-12}+\\sqrt{18x-27}\\,$$=12-\\sqrt{2x-3} $<br\/>$ \\Leftrightarrow 2\\sqrt{2x-3}+3\\sqrt{2x-3}\\,$$+\\sqrt{2x-3}=12 $<br\/>$ \\Leftrightarrow 6\\sqrt{2x-3}=12 $<br\/>$ \\Leftrightarrow \\sqrt{2x-3}\\,\\,\\,=2 $<br\/>$\\Leftrightarrow 2x-3\\,\\,\\,\\,\\,\\,\\,\\,\\,=4 $<br\/>$\\Leftrightarrow x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\dfrac{7}{2}\\,\\,\\,\\,\\,\\text{(th\u1ecfa m\u00e3n)} $ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ \\dfrac{7}{2} \\right\\}$<\/span>"}]}],"id_ques":745},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $1-2\\sqrt{2}$","B. $1+2\\sqrt{2}$","C. $-2\\sqrt{2}$"],"ques":"T\u00ednh:<br\/>$\\dfrac{1}{\\sqrt{2}-1}-\\dfrac{3\\sqrt{6}-3\\sqrt{10}}{\\sqrt{3}-\\sqrt{5}}=$?","hint":"Tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu c\u1ee7a ph\u00e2n th\u1ee9c th\u1ee9 nh\u1ea5t v\u00e0 t\u00ecm th\u1eeba s\u1ed1 chung \u1edf ph\u00e2n th\u1ee9c th\u1ee9 2","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{align} & \\,\\,\\,\\,\\,\\dfrac{1}{\\sqrt{2}-1}-\\dfrac{3\\sqrt{6}-3\\sqrt{10}}{\\sqrt{3}-\\sqrt{5}} \\\\ & =\\dfrac{\\sqrt{2}+1}{(\\sqrt{2})^2-1}-\\dfrac{3\\sqrt{2}\\left( \\sqrt{3}-\\sqrt{5} \\right)}{\\sqrt{3}-\\sqrt{5}} \\\\ & =\\sqrt{2}+1-3\\sqrt{2} \\\\ & =1-2\\sqrt{2} \\\\ \\end{align}$<\/span>"}]}],"id_ques":746},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank_random","correct":[[["<"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai7/lv2/img\/5.jpg' \/><\/center> Cho hai bi\u1ec3u th\u1ee9c $A=\\sqrt{11}-\\sqrt{10}$ v\u00e0 $B=\\sqrt{4}-\\sqrt{3}$ <br\/>So s\u00e1nh $A$ _input_$B$","hint":"$A=\\sqrt{11}-\\sqrt{10}=\\dfrac{1}{\\sqrt{11}+\\sqrt{10}}$<br\/>$B=\\sqrt{4}-\\sqrt{3}=\\dfrac{1}{\\sqrt{4}+\\sqrt{3}}$ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: Nh\u00e2n c\u1ea3 t\u1eed v\u00e0 m\u1eabu c\u1ee7a $A,B$ v\u1edbi bi\u1ec3u th\u1ee9c li\u00ean h\u1ee3p.<br\/>B\u01b0\u1edbc 2: So s\u00e1nh $A$ v\u00e0 $B$. <br\/>\u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t: N\u1ebfu $a>b$ th\u00ec $\\dfrac{1}{a} < b$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/> $A=\\sqrt{11}-\\sqrt{10}\\,$$=\\dfrac{\\left( \\sqrt{11}-\\sqrt{10} \\right)\\left( \\sqrt{11}+\\sqrt{10} \\right)}{\\left( \\sqrt{11}+\\sqrt{10} \\right)}\\,$$=\\dfrac{1}{\\sqrt{11}+\\sqrt{10}}$<br\/>$B=\\sqrt{4}-\\sqrt{3}\\,$$=\\dfrac{\\left( \\sqrt{4}-\\sqrt{3} \\right)\\left( \\sqrt{4}+\\sqrt{3} \\right)}{\\left( \\sqrt{4}+\\sqrt{3} \\right)}\\,$$=\\dfrac{1}{\\sqrt{4}+\\sqrt{3}}$<br\/>M\u00e0: $\\sqrt{4}<\\sqrt{11};\\,\\,\\,\\sqrt{3}<\\sqrt{10}$$\\Rightarrow \\sqrt{11}+\\sqrt{10}>\\sqrt{4}+\\sqrt{3}$ <br\/>Do \u0111\u00f3 $\\dfrac{1}{\\sqrt{11}+\\sqrt{10}}<\\dfrac{1}{\\sqrt{4}+\\sqrt{3}}\\,$$\\Leftrightarrow \\sqrt{11}-\\sqrt{10}<\\sqrt{4}-\\sqrt{3}$ <br\/><span class='basic_pink'>V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0 $<$ <\/span><\/span><\/span>"}]}],"id_ques":747},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Bi\u1ec3u th\u1ee9c $2\\sqrt{x+5}+x\\sqrt{4-x}$ c\u00f3 ngh\u0129a khi: ","select":["A. $x \\ge 4$ ","B. $x \\le -5$","C. $-5\\le x\\le 4$","D. $x \\ge 4$ ho\u1eb7c $x \\le - 5$"],"hint":" $\\sqrt{A}$ c\u00f3 ngh\u0129a $\\Leftrightarrow A \\ge 0$ ","explain":"<span class='basic_left'> Bi\u1ec3u th\u1ee9c c\u00f3 ngh\u0129a khi:<br\/>$\\left\\{ \\begin{aligned} & x+5\\ge 0 \\\\ & 4-x\\ge 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\ge -5 \\\\ & x\\le 4 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow -5\\le x\\le 4$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span>","column":2}]}],"id_ques":748},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai7/lv2/img\/2.jpg' \/><\/center>K\u1ebft qu\u1ea3 ph\u00e9p t\u00ednh $\\sqrt{7+\\sqrt{24}}.\\sqrt{7-2\\sqrt{6}}$ l\u00e0: ","select":["A. $\\sqrt{49+7\\sqrt{24}-14\\sqrt{6}}$ ","B. $25$","C. $5$","D. $\\sqrt{5}$"],"hint":": \u00c1p d\u1ee5ng quy t\u1eafc nh\u00e2n hai c\u0103n th\u1ee9c v\u00e0 s\u1eed d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{align} & \\,\\,\\,\\,\\,\\,\\sqrt{7+\\sqrt{24}}.\\sqrt{7-2\\sqrt{6}} \\\\ & =\\sqrt{7+2\\sqrt{6}}.\\sqrt{7-2\\sqrt{6}} \\\\ & =\\sqrt{\\left( 7+2\\sqrt{6} \\right)\\left( 7-2\\sqrt{6} \\right)} \\\\ & =\\sqrt{{{7}^{2}}-{{\\left( 2\\sqrt{6} \\right)}^{2}}} \\\\ & =\\sqrt{25} \\\\ & =5 \\\\ \\end{align}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span>","column":2}]}],"id_ques":749},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai7/lv2/img\/4.jpg' \/><\/center>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $\\sqrt{2x-1}+\\dfrac{x-1}{x-6}$ l\u00e0:","select":["A. $x\\ge \\dfrac{1}{2}$ v\u00e0 $x \\ne 6$ ","B. $x\\ge -\\dfrac{1}{2}$ v\u00e0 $x \\ne 6$","C. $x\\ge \\dfrac{1}{2}$ v\u00e0 $x > 6$","D. $x\\ge \\dfrac{1}{2}$ v\u00e0 $x \\ne 1$"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: <br\/>$\\left\\{ \\begin{aligned} & 2x-1\\ge 0 \\\\ & x-6\\ne 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\ge \\dfrac{1}{2} \\\\ & x\\ne 6 \\\\ \\end{aligned} \\right.$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span>","column":2}]}],"id_ques":750}],"lesson":{"save":0,"level":2}}