{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":" K\u1ebft qu\u1ea3 ph\u00e2n t\u00edch bi\u1ec3u th\u1ee9c $x-\\sqrt{xy}+4\\sqrt{x}-2\\sqrt{y}+4$ th\u00e0nh nh\u00e2n t\u1eed l\u00e0: ","select":["A. $\\left( \\sqrt{x}-2 \\right)\\left( \\sqrt{x}+2-\\sqrt{y} \\right)$","B. $\\left( \\sqrt{x}+2 \\right)\\left( \\sqrt{x}+2 -\\sqrt{y}\\right)$","C. $\\left( \\sqrt{x}+2 \\right)\\left( \\sqrt{x}+2 +\\sqrt{y}\\right)$","D. $\\left( \\sqrt{x}-2 \\right)\\left( \\sqrt{x}+2 +\\sqrt{y}\\right)$ "],"hint":"Ph\u00e2n t\u00edch \u0111a th\u1ee9c b\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p nh\u00f3m h\u1ea1ng t\u1eed","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span><b>B\u01b0\u1edbc 1:<\/b> Nh\u00f3m c\u00e1c h\u1ea1ng t\u1eed m\u1ed9t c\u00e1ch h\u1ee3p l\u00fd l\u00e0m xu\u1ea5t hi\u1ec7n nh\u00e2n t\u1eed chung <br\/><b>B\u01b0\u1edbc 2:<\/b> \u0110\u1eb7t nh\u00e2n t\u1eed chung <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/>$\\begin{align} & \\,\\,\\,x-\\sqrt{xy}+4\\sqrt{x}-2\\sqrt{y}+4 \\\\ & =\\left( x+4\\sqrt{x}+4 \\right)-\\left( \\sqrt{xy}+2\\sqrt{y} \\right) \\\\ & ={{\\left( \\sqrt{x}+2 \\right)}^{2}}-\\sqrt{y}\\left( \\sqrt{x}+2 \\right) \\\\ & =\\left( \\sqrt{x}+2 \\right)\\left( \\sqrt{x}+2-\\sqrt{y} \\right) \\\\ \\end{align}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B <\/span> <\/span>","column":2}]}],"id_ques":601},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["0"],["9"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai4/lv3/img\/2.jpg' \/><\/center><span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $\\,\\,P=\\dfrac{\\sqrt{x}+1}{\\sqrt{x}-3}$ <br\/><b> C\u00e2u 1: <\/b> \u0110i\u1ec1u ki\u1ec7n \u0111\u1ec3 $P$ c\u00f3 ngh\u0129a l\u00e0 $x\\ge$_input_ v\u00e0 $x\\ne$_input_<\/span>","hint":"\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c: Bi\u1ec3u th\u1ee9c trong c\u0103n ph\u1ea3i kh\u00f4ng \u00e2m v\u00e0 m\u1eabu ph\u1ea3i kh\u00e1c $0$","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: <br\/> $\\left\\{ \\begin{aligned} & x\\ge 0 \\\\ & \\sqrt{x}-3\\ne 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\ge 0 \\\\ & x\\ne 9 \\\\ \\end{aligned} \\right.$ <br\/><span class='basic_pink'>V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0$ v\u00e0 $9$<\/span><\/span>"}]}],"id_ques":602},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank_random","correct":[[["49"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $\\,\\,P=\\dfrac{\\sqrt{x}+1}{\\sqrt{x}-3}$ <br\/><b> C\u00e2u 2: <\/b>Gi\u00e1 tr\u1ecb nguy\u00ean l\u1edbn nh\u1ea5t c\u1ee7a $x$ \u0111\u1ec3 $P \\in \\mathbb Z$ l\u00e0 $x=$_input_<\/span>","hint":"Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c v\u1ec1 d\u1ea1ng $m+\\dfrac {n}{f(x)}$, trong \u0111\u00f3 $m,n \\in \\mathbb Z$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c.<br\/><b>B\u01b0\u1edbc 2:<\/b> Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c v\u1ec1 d\u1ea1ng $m+\\dfrac {n}{f(x)}$, trong \u0111\u00f3 $m,n \\in \\mathbb Z$<br\/><b>B\u01b0\u1edbc 3:<\/b> T\u00ecm $x$ \u0111\u1ec3 $P$ nguy\u00ean. <br\/><b>B\u01b0\u1edbc 4:<\/b> T\u00ecm $x$ nguy\u00ean l\u1edbn nh\u1ea5t v\u00e0 k\u1ebft lu\u1eadn.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'>V\u1edbi $x\\ge 0$ v\u00e0 $x \\ne 9$, ta c\u00f3: <br\/>Ta c\u00f3:<br\/>$P=\\dfrac{\\sqrt{x}+1}{\\sqrt{x}-3}=\\dfrac{\\sqrt{x}-3+3+1}{\\sqrt{x}-3}\\,$$=1+\\dfrac{4}{\\sqrt{x}-3}$<br\/>\u0110\u1ec3 $P$ nguy\u00ean th\u00ec $\\dfrac{4}{\\sqrt{x}-3}$ ph\u1ea3i c\u00f3 gi\u00e1 tr\u1ecb nguy\u00ean.<br\/> V\u1eady $\\sqrt{x}-3\\in \u01af(4)=\\{\\pm 1; \\pm2; \\pm 4\\}$ <br\/>Ta c\u00f3 b\u1ea3ng sau:<br\/><table> <tr> <th>$\\sqrt{x}-3$<\/th> <th>$-4$<\/th> <th>$-2$<\/th> <th>$-1$ <\/th> <th>$1$ <\/th> <th>$2$<\/th> <th>$4$<\/th> <\/tr> <tr> <td>$\\sqrt{x}$<\/td> <td>-1<\/td> <td>$1$<\/td> <td>$2$<\/td> <td>$4$<\/td> <td>$5$<\/td> <td>$7$<\/td> <\/tr> <tr> <td>$x$<\/td> <td> lo\u1ea1i <\/td> <td>$1$<\/td> <td>$4$<\/td> <td>$16$<\/td> <td>$25$<\/td> <td>$49$<\/td> <\/tr><\/table><br\/>V\u1eady gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a $x$ t\u00ecm \u0111\u01b0\u1ee3c \u0111\u1ec3 $P$ nh\u1eadn gi\u00e1 tr\u1ecb nguy\u00ean l\u00e0 $x=49$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $49$<\/span><\/span><\/span>"}]}],"id_ques":603},{"time":24,"part":[{"title":"\u0110i\u1ec1n bi\u1ec3u th\u1ee9c th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["1-a","-a+1"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"V\u1edbi $a\\ge 0;\\,\\,\\,a\\ne 1$ . R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c:<br\/> $\\left( 1+\\dfrac{a+\\sqrt{a}}{1+\\sqrt{a}} \\right)\\left( 1-\\dfrac{a-\\sqrt{a}}{\\sqrt{a}-1} \\right)=$_input_ ","hint":"","explain":"<span class='basic_left'>V\u1edbi $a\\ge 0;\\,\\,\\,a\\ne 1$, ta c\u00f3:<br\/>$ \\,\\,\\left( 1+\\dfrac{a+\\sqrt{a}}{1+\\sqrt{a}} \\right)\\left( 1-\\dfrac{a-\\sqrt{a}}{\\sqrt{a}-1} \\right) $<br\/>$ =\\left[ 1+\\dfrac{\\sqrt{a}\\left( \\sqrt{a}+1 \\right)}{\\sqrt{a}+1} \\right]$$\\left[ 1-\\dfrac{\\sqrt{a}\\left( \\sqrt{a}-1 \\right)}{\\sqrt{a}-1} \\right]$<br\/>$=\\left( 1+\\sqrt{a} \\right)\\left( 1-\\sqrt{a} \\right) $<br\/>$=1-a $ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1-a$ <\/span><\/span>"}]}],"id_ques":604},{"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":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai4/lv3/img\/5.jpg' \/><\/center>$\\sqrt{15}-\\sqrt{13}$ _input_ $\\sqrt{13}-\\sqrt{11}$ ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> Bi\u1ebfn \u0111\u1ed5i v\u1ec1 so s\u00e1nh $\\sqrt{15}+\\sqrt{13}$ v\u00e0 $\\sqrt{13}+\\sqrt{11}$ <br\/><b>B\u01b0\u1edbc 2:<\/b> So s\u00e1nh<\/span> <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'>Ta c\u00f3: $\\sqrt{15}-\\sqrt{13}\\,$$=\\dfrac{\\left( \\sqrt{15}-\\sqrt{13} \\right)\\left( \\sqrt{15}+\\sqrt{13} \\right)}{\\sqrt{15}+\\sqrt{13}}\\,$$=\\dfrac{(\\sqrt{15})^{2}-(\\sqrt{13})^{2}}{\\sqrt{15}+\\sqrt{13}}\\,$$=\\dfrac{2}{\\sqrt{15}+\\sqrt{13}}$<br\/>$\\sqrt{13}-\\sqrt{11}\\,$$=\\dfrac{\\left( \\sqrt{13}-\\sqrt{11} \\right)\\left( \\sqrt{13}+\\sqrt{11} \\right)}{\\sqrt{13}+\\sqrt{11}}\\,$$=\\dfrac{2}{\\sqrt{13}+\\sqrt{11}}$<br\/>V\u00ec $\\sqrt{15}+\\sqrt{13}>\\sqrt{13}+\\sqrt{11}\\,$$\\Rightarrow \\dfrac{2}{\\sqrt{15}+\\sqrt{13}}<\\dfrac{2}{\\sqrt{13}+\\sqrt{11}}\\,$$\\Rightarrow \\sqrt{15}-\\sqrt{13}<\\sqrt{13}-\\sqrt{11}$ <\/span> <br\/><span class='basic_pink'>V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0 $<$ <\/span> <\/span><\/span>"}]}],"id_ques":605},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $a-\\sqrt{a}$","B. $a+\\sqrt{a}$","C. $2a -\\sqrt{a}$"],"ques":"<span class='basic_left'>Cho bi\u1ec3u th\u1ee9c: $M=\\dfrac{{{a}^{2}}+\\sqrt{a}}{a-\\sqrt{a}+1}-\\dfrac{2a+\\sqrt{a}}{\\sqrt{a}}+1$<br\/>R\u00fat g\u1ecdn $M=$?<\/span> ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> \u0110\u1eb7t \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 bi\u1ec3u th\u1ee9c c\u00f3 ngh\u0129a<br\/><b>B\u01b0\u1edbc 2:<\/b>R\u00fat g\u1ecdn t\u1eebng ph\u00e2n th\u1ee9c trong bi\u1ec3u th\u1ee9c b\u1eb1ng c\u00e1ch ph\u00e2n t\u00edch t\u1eed th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed.<br\/><b>B\u01b0\u1edbc 3:<\/b> S\u1eed d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $A^3 \\pm B^3$<br\/><b>B\u01b0\u1edbc 4:<\/b> R\u00fat g\u1ecdn <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> \u0110i\u1ec1u ki\u1ec7n: $a > 0$<br\/>Ta c\u00f3:<br\/>$M=\\dfrac{{{a}^{2}}+\\sqrt{a}}{a-\\sqrt{a}+1}-\\dfrac{2a+\\sqrt{a}}{\\sqrt{a}}+1$ <br\/> $=\\dfrac{(\\sqrt{a})^{4}+\\sqrt{a}}{a-\\sqrt{a}+1}\\,$$-\\dfrac{2(\\sqrt{a})^{2}+\\sqrt{a}}{\\sqrt{a}}+1$<br\/>$=\\dfrac{\\sqrt{a}\\left[(\\sqrt{a})^{3}+1 \\right]}{a-\\sqrt{a}+1}\\,$$-\\dfrac{\\sqrt{a}\\left( 2\\sqrt{a}+1 \\right)}{\\sqrt{a}}+1 $<br\/>$=\\sqrt{a}\\left( \\sqrt{a}+1 \\right)-\\left( 2\\sqrt{a}+1 \\right)+1 $<br\/>$=a+\\sqrt{a}-2\\sqrt{a}-1+1$<br\/>$=a-\\sqrt{a}$<\/span>"}]}],"id_ques":606},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["="]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'>Cho bi\u1ec3u th\u1ee9c: $M=\\dfrac{{{a}^{2}}+\\sqrt{a}}{a-\\sqrt{a}+1}-\\dfrac{2a+\\sqrt{a}}{\\sqrt{a}}+1$<br\/><b> C\u00e2u 2:<\/b> V\u1edbi $a > 1$, so s\u00e1nh $M$ v\u00e0 $|M|$<br\/><b>\u0110\u00e1p \u00e1n:<\/b> $M$ _input_ $|M|$<\/span>","hint":"So s\u00e1nh $M$ v\u1edbi $0$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> Ch\u1ee9ng minh $M > 0$ <br\/><b>B\u01b0\u1edbc 2:<\/b> \u00c1p d\u1ee5ng: $\\left| a \\right|=\\left\\{ \\begin{align} & \\begin{matrix} \\,\\,\\,a & n\u1ebfu\\,\\,\\, a\\ge0 \\\\\\end{matrix} \\\\ & \\begin{matrix} -a & n\u1ebfu \\,\\,\\,a < 0 \\\\\\end{matrix} \\\\ \\end{align} \\right.$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Theo c\u00e2u tr\u00ean, ta c\u00f3: $M=a-\\sqrt{a}$<br\/>Ta c\u00f3:<br\/> $M=a-\\sqrt{a}=\\sqrt{a}\\left( \\sqrt{a}-1 \\right)$<br\/>M\u00e0 $a > 1 \\Rightarrow \\sqrt{a}-1>0$<br\/>$\\Rightarrow \\sqrt{a}\\left( \\sqrt{a}-1 \\right)>0\\Rightarrow M>0$<br\/>Do \u0111\u00f3 $|M|=M$<br\/><span class='basic_pink'>V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $=$<\/span><\/span>"}]}],"id_ques":607},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["4"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'>Cho bi\u1ec3u th\u1ee9c: $\\,\\,\\,M=\\dfrac{{{a}^{2}}+\\sqrt{a}}{a-\\sqrt{a}+1}-\\dfrac{2a+\\sqrt{a}}{\\sqrt{a}}+1$<br\/><b> C\u00e2u 3: <\/b>\u0110\u1ec3 $M= 2$ th\u00ec $a =$ _input_ ","hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $M=2$ \u0111\u1ec3 t\u00ecm $a$<\/span>","explain":"<span class='basic_left'> V\u1edbi $a>0$, $M=a-\\sqrt{a}$<br\/>Ta c\u00f3: <br\/>$M= 2$<br\/>$ \\Leftrightarrow a-\\sqrt{a}=2$<br\/>$\\Leftrightarrow a-\\sqrt{a}-2=0$<br\/> $\\Leftrightarrow a+\\sqrt{a} - 2\\sqrt{a}-2=0$<br\/> $\\Leftrightarrow \\sqrt{a} (\\sqrt{a} + 1) - 2(\\sqrt{a} + 1)=0$<br\/>$\\Leftrightarrow \\left( \\sqrt{a}+1 \\right)\\left( \\sqrt{a}-2 \\right)=0$<br\/>$\\Leftrightarrow \\left[ \\begin{aligned} & \\sqrt{a}=-1\\,\\,\\,\\,\\,\\, \\text{(lo\u1ea1i)} \\\\ & \\sqrt{a}=2 \\\\ \\end{aligned} \\right.$ <br\/> $\\Leftrightarrow a=4$ (th\u1ecfa m\u00e3n)<br\/>Do \u0111\u00f3 v\u1edbi $M = 2$ th\u00ec $a=4$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $4$ <\/span><\/span>"}]}],"id_ques":608},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"select":["A. $M_{min} =-\\dfrac{1}{3}$ khi $a = \\dfrac{1}{4}$","B. $M_{min} =-\\dfrac{1}{4}$ khi $a = \\dfrac{1}{4}$","C. $M_{min} =-\\dfrac{1}{4}$ khi $a = \\dfrac{1}{2}$"],"ques":"<span class='basic_left'>Cho bi\u1ec3u th\u1ee9c: $M=\\dfrac{{{a}^{2}}+\\sqrt{a}}{a-\\sqrt{a}+1}-\\dfrac{2a+\\sqrt{a}}{\\sqrt{a}}+1$<br\/><b> C\u00e2u 4: <\/b>Gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $M=$?khi $a=$?<\/span>","hint":"Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c $M$ v\u1ec1 d\u1ea1ng: ${f(x)}^2+a$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> \u0110\u01b0a bi\u1ec3u th\u1ee9c v\u1ec1 d\u1ea1ng ${{\\left[ f\\left( x \\right) \\right]}^{2}}+a$<br\/><b>B\u01b0\u1edbc 2:<\/b> \u0110\u00e1nh gi\u00e1 v\u00e0 x\u00e1c \u0111\u1ecbnh gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $M$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'>V\u1edbi $a>0$. Ta c\u00f3: <br\/>$M=a-\\sqrt{a}\\,$ <br\/> $={{\\left( \\sqrt{a} \\right)}^{2}}-\\sqrt{a}$ <br\/> $ ={{\\left( \\sqrt{a} \\right)}^{2}}-2.\\sqrt{a}.\\dfrac{1}{2}+{{\\left( \\dfrac{1}{2} \\right)}^{2}}-{{\\left( \\dfrac{1}{2} \\right)}^{2}}$ <br\/> $={{\\left( \\sqrt{a}-\\dfrac{1}{2} \\right)}^{2}}-\\dfrac{1}{4}\\ge -\\dfrac{1}{4}$ v\u1edbi m\u1ecdi $a$ <br\/>$\\Rightarrow{{M}_{\\min }}=-\\dfrac{1}{4}\\Leftrightarrow \\sqrt{a}-\\dfrac{1}{2}=0\\,$$\\Leftrightarrow \\sqrt{a}=\\dfrac{1}{2}\\Leftrightarrow a=\\dfrac{1}{4}$<br\/><\/span><\/span>"}]}],"id_ques":609},{"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","-2"],["-2","2"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh:<br\/> $\\sqrt{9{{x}^{2}}+45}-\\dfrac{1}{12}\\sqrt{16{{x}^{2}}+80}\\,$$+3\\sqrt{\\dfrac{{{x}^{2}}+5}{16}}\\,$$-\\dfrac{1}{4}\\sqrt{\\dfrac{25{{x}^{2}}+125}{9}}=9$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${_input_ ;_input_} ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 ngh\u0129a.<br\/><b>B\u01b0\u1edbc 2:<\/b> \u00c1p d\u1ee5ng: Quy t\u1eafc \u0111\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n: $\\sqrt{A^2B}=|A|\\sqrt{B}$<br\/><b>B\u01b0\u1edbc 3:<\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c <br\/><b>B\u01b0\u1edbc 4:<\/b> B\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf r\u1ed3i t\u00ecm nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $x\\in \\mathbb{R}$ <br\/> Ta c\u00f3:<br\/> $\\sqrt{9{{x}^{2}}+45}-\\dfrac{1}{12}\\sqrt{16{{x}^{2}}+80}\\,$$+3\\sqrt{\\dfrac{{{x}^{2}}+5}{16}}\\,$$-\\dfrac{1}{4}\\sqrt{\\dfrac{25{{x}^{2}}+125}{9}}=9$<br\/>$\\Leftrightarrow \\sqrt{9({{x}^{2}}+5)}-\\dfrac{1}{12}\\sqrt{16({{x}^{2}}+5)}\\,$$+\\dfrac{3}{4}\\sqrt{{{x}^{2}}+5}\\,$$-\\dfrac{1}{4}.\\dfrac{1}{3}\\sqrt{25({{x}^{2}}+5)}=9$<br\/>$\\Leftrightarrow 3\\sqrt{{{x}^{2}}+5}-\\dfrac{1}{3}\\sqrt{{{x}^{2}}+5}\\,$$+\\dfrac{3}{4}\\sqrt{{{x}^{2}}+5}\\,$$-\\dfrac{5}{12}\\sqrt{{{x}^{2}}+5}=9$<br\/>$\\begin{aligned} & \\Leftrightarrow 3\\sqrt{{{x}^{2}}+5}=9 \\\\ & \\Leftrightarrow \\sqrt{{{x}^{2}}+5}=3 \\\\ & \\Leftrightarrow {{x}^{2}}+5=9 \\\\ & \\Leftrightarrow {{x}^{2}}=4 \\\\ & \\Leftrightarrow x=\\pm 2 \\\\ \\end{aligned}$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{2; -2\\}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2$ v\u00e0 $-2$ <\/span> <\/span><\/span>"}]}],"id_ques":610}],"lesson":{"save":0,"level":3}}