{"segment":[{"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"K\u1ebft qu\u1ea3 ph\u00e9p nh\u00e2n $2{{x}^{3}}\\left( 3{{x}^{2}}-x+\\dfrac{1}{2} \\right)$ l\u00e0:","select":["A. $6x^5-2x^4+x^3$","B. $5x^5-2x^4+4x^3$","C. $6x^6-2x^3+x^3$ ","D. $6x^5-2x^3+4x^2$"],"hint":" Th\u1ef1c hi\u1ec7n ph\u00e9p nh\u00e2n \u0111\u01a1n th\u1ee9c v\u1edbi \u0111a th\u1ee9c: $A(B+C)=A.B+A.C$","explain":"<span class='basic_left'>Ta c\u00f3: <br\/> $2{{x}^{3}}\\left( 3{{x}^{2}}-x+\\dfrac{1}{2} \\right)=6{{x}^{5}}-2{{x}^{4}}+{{x}^{3}}$ <br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n l\u00e0 A.<\/span>","column":2}],"id_ques":781},{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":" K\u1ebft qu\u1ea3 khai tri\u1ec3n $x^3-125$ l\u00e0:","select":["A. $(x-5)(x+5)$","B. $(x-5)(x^2+5x+25)$","C. $(x-15)(x+15)$ ","D. $(x-5)(x^2-5x+25)$"],"hint":" Khai tri\u1ec3n theo h\u1eb1ng \u0111\u1eb3ng th\u1ee9c: $a^3-b^3=(a-b)(a^2+ab+b^2)$","explain":"<span class='basic_left'> $\\begin{align} & x^3-125 \\\\ & =x^3-5^3 \\\\ & =(x-5)(x^2+5x+25) \\\\ \\end{align}$ <br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n l\u00e0 B.<\/span>","column":2}],"id_ques":782},{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":" Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $125x^3-75x^2+15x-1$ t\u1ea1i $x = \\dfrac{1}{5}$ l\u00e0:","select":["A. $125$","B. $27$","C. $-1$ ","D. $0$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1:<\/b> S\u1eed d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $(a-b)^3=a^3-3a^2b+3ab^2-b^3$ \u0111\u1ec3 thu g\u1ecdn bi\u1ec3u th\u1ee9c .<br\/> <b> B\u01b0\u1edbc 2:<\/b> Thay $x = \\dfrac{1}{5}$ v\u00e0o \u0111\u1ec3 t\u00ednh.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta c\u00f3: <br\/> $125x^3-75x^2+15x-1= (5x)^3 - 3 . (5x)^2 + 3 . 5x - 1^3 = (5x-1)^3$.<br\/> Thay $x = \\dfrac{1}{5}$ v\u00e0o bi\u1ec3u th\u1ee9c $(5x-1)^3$, ta \u0111\u01b0\u1ee3c: $(5.\\dfrac{1}{5}-1)^3=0$. <br\/> <span class='basic_pink'> V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n D. <\/span><\/span> ","column":2}],"id_ques":783},{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":" K\u1ebft qu\u1ea3 ph\u00e2n t\u00edch \u0111a th\u1ee9c $4x^2-4xy+y^2-9$ th\u00e0nh nh\u00e2n t\u1eed l\u00e0","select":["A. $(2x-y)(2x+y)$","B. $(2x+y+3)(2x-y-3)$","C. $(2x-y+3)(2x-y-3)$ ","D. $4(2x+y)(2x-y)$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1:<\/b> Nh\u00f3m c\u00e1c h\u1ea1ng t\u1eed m\u1ed9t c\u00e1ch th\u00edch h\u1ee3p l\u00e0m xu\u1ea5t hi\u1ec7n h\u1eb1ng \u0111\u1eb3ng th\u1ee9c hi\u1ec7u hai b\u00ecnh ph\u01b0\u01a1ng <br\/> <b> B\u01b0\u1edbc 2:<\/b> Ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed theo h\u1eb1ng \u0111\u1eb3ng th\u1ee9c hi\u1ec7u hai b\u00ecnh ph\u01b0\u01a1ng <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta c\u00f3:<br\/> $\\begin{align} & 4x^2-4xy+y^2-9\\\\ & =(4x^2-4xy+y^2)-9\\\\ & =(2x-y)^2-3^2 \\\\ & =(2x-y+3)(2x-y-3) \\\\ \\end{align}$. <br\/> <span class='basic_pink'> V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n C. <\/span><\/span> ","column":2}],"id_ques":784},{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng trong ph\u00e9p chia sau","title_trans":"","temp":"fill_the_blank","correct":[[["4x-11","-11+4x"],["26x-10","-10+26x"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai11/lv3/img\/2.jpg' \/><\/center> $(4x^3-3x^2+1):(x^2+2x-1)=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ d\u01b0 $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$","hint":" \u0110\u1eb7t t\u00ednh chia sau \u0111\u00f3 t\u00ecm th\u01b0\u01a1ng v\u00e0 t\u00ecm s\u1ed1 d\u01b0.","explain":"<span class='basic_left'> \u0110\u1eb7t ph\u00e9p chia: <br\/> $\\left. \\begin{align} & \\begin{matrix} 4{{x}^{3}}-3{{x}^{2}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+1\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\ 4{{x}^{3}}+8{{x}^{2}}-4x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\end{matrix} \\\\ & \\overline{\\begin{align} & \\,\\,\\,\\,\\,\\,\\,\\,\\,-11{{x}^{2}}+4x+1\\,\\,\\, \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,-11{{x}^{2}}-22x+11\\,\\,\\,\\, \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\overline{\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,26x-10\\,\\,\\,\\,} \\\\ \\end{align}} \\\\ \\end{align} \\right|\\begin{matrix} \\dfrac{{{x}^{2}}+2x-1}{4x-11} \\\\ \\begin{matrix} {} \\\\ {} \\\\\\end{matrix} \\\\ {} \\\\\\end{matrix}$ <br\/> V\u1eady $(4x^3-3x^2+1):(x^2+2x-1)=4x-11$ d\u01b0 $26x-10$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 \u0111\u00e1p \u00e1n ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $4x-11$ v\u00e0 $26x-10$. <\/span><\/span> "}],"id_ques":785},{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng trong ph\u00e9p chia sau","title_trans":"","temp":"fill_the_blank","correct":[[["0"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai11/lv3/img\/4.jpg' \/><\/center> S\u1ed1 d\u01b0 c\u1ee7a ph\u00e9p chia $(2x^4-3x^3-3x^2+6x-2):(x^2-2)$ l\u00e0 _input_","hint":" \u0110\u1eb7t ph\u00e9p chia $(2x^4-3x^3-3x^2+6x-2):(x^2-2)$, t\u00ecm s\u1ed1 d\u01b0. ","explain":"<span class='basic_left'> \u0110\u1eb7t ph\u00e9p chia: <br\/> $\\left. \\begin{align} & \\begin{matrix} 2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2\\,\\,\\,\\, \\\\ 2{{x}^{4}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-4{{x}^{2}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\end{matrix} \\\\ & \\overline{\\begin{align} & \\,\\,\\,\\,\\,\\,\\,\\,\\,-3{{x}^{3}}+{{x}^{2}}+6x-2\\,\\,\\,\\,\\,\\,\\, \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,-3{{x}^{3}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+6x\\,\\,\\,\\, \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\overline{\\begin{align} & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{{x}^{2}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-2\\,\\,\\,\\,\\,\\,\\,\\, \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{{x}^{2}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-2\\,\\,\\,\\,\\,\\,\\,\\, \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\overline{\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,} \\\\ \\end{align}} \\\\ \\end{align}} \\\\ \\end{align} \\right|\\begin{matrix} \\dfrac{{{x}^{2}}-2}{2{{x}^{2}}-3x+1} \\\\ \\begin{matrix} \\begin{matrix} \\begin{matrix} {} \\\\ {} \\\\\\end{matrix} \\\\ {} \\\\\\end{matrix} \\\\ {} \\\\\\end{matrix} \\\\ {} \\\\\\end{matrix}$<br\/> S\u1ed1 d\u01b0 c\u1ee7a ph\u00e9p chia tr\u00ean l\u00e0 $0$. <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0$. <\/span><\/span> "}],"id_ques":786},{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["15"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai11/lv3/img\/10.jpg' \/><\/center> T\u00ecm $m$ $( m\\,\\in\\,\\mathbb{Z})$ \u0111\u1ec3 ph\u00e9p chia $(3x^2+mx+27):(x+5)$ c\u00f3 s\u1ed1 d\u01b0 l\u00e0 $27$ <br\/> \u0110\u00e1p \u00e1n: $m$ = _input_","hint":" \u0110\u1eb7t t\u00ednh chia r\u1ed3i t\u00ecm s\u1ed1 d\u01b0, cho s\u1ed1 d\u01b0 b\u1eb1ng $27$ \u0111\u1ec3 t\u00ecm $m$.","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\left. \\begin{align} & \\begin{matrix} 3{{x}^{2}}+mx+27\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\ 3{{x}^{2}}+15x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\end{matrix} \\\\ & \\overline{\\begin{align} & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( m-15 \\right)x+27\\,\\,\\,\\,\\,\\,\\, \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( m-15 \\right)x+5m-75\\,\\,\\,\\, \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\overline{\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,102-5m\\,\\,\\,\\,\\,\\,\\,} \\\\ \\end{align}} \\\\ \\end{align} \\right|\\begin{matrix} \\dfrac{x+5}{3x+m-15} \\\\ \\begin{matrix} {} \\\\ {} \\\\\\end{matrix} \\\\ {} \\\\\\end{matrix}$ <br\/> V\u1eady $(3x^2+mx+27):(x+5)=3x+m-15$ d\u01b0 $102-5m$<br\/> \u0110\u1ec3 $(3x^2+mx+27):(x+5)$ c\u00f3 d\u01b0 l\u00e0 $27$ th\u00ec <br\/> $102-5m=27\\Rightarrow m=15$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 \u0111\u00e1p \u00e1n ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $15$. <\/span><\/span> "}],"id_ques":787},{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai11/lv3/img\/11.jpg' \/><\/center> Gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c $x^2+y^2+z^2+4x-2y-4z+10$ l\u00e0 _input_","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 $x^2+y^2+z^2+4x-2y-4z+10$ v\u1ec1 d\u1ea1ng $[f(x)]^2+[f(y)]^2+[f(z)]^2+a$ v\u1edbi $a$ l\u00e0 h\u1eb1ng s\u1ed1. <br\/> <b> B\u01b0\u1edbc 2:<\/b> Gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c l\u00e0 $a$, t\u00ecm $x, y, z$ \u0111\u1ec3 d\u1ea5u \"=\" x\u1ea3y ra.<\/b> <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> $\\begin{aligned} & {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+4x-2y-4z+10 \\\\ & =\\left( {{x}^{2}}+4x+4 \\right)+\\left( {{y}^{2}}-2y+1 \\right)+\\left( {{z}^{2}}-4z+4 \\right)+1 \\\\ & ={{\\left( x+2 \\right)}^{2}}+{{\\left( y-1 \\right)}^{2}}+{{\\left( z-2 \\right)}^{2}}+1 \\\\ & \\text{Do}\\,\\,\\,\\left. \\begin{aligned} & {{\\left( x+2 \\right)}^{2}}\\ge 0 \\\\ & {{\\left( y-1 \\right)}^{2}}\\,\\ge 0 \\\\ & {{\\left( z-2 \\right)}^{2}}\\ge 0 \\\\ \\end{aligned} \\right\\}\\Rightarrow {{\\left( x+2 \\right)}^{2}}+{{\\left( y-1 \\right)}^{2}}+{{\\left( z-2 \\right)}^{2}}+1\\ge 1 \\hspace{0,2cm} \\text{v\u1edbi} \\hspace{0,2cm} \\forall x \\\\ \\end{aligned}$ <br\/> V\u1eady GTNN c\u1ee7a bi\u1ec3u th\u1ee9c l\u00e0 $1$.<br\/> D\u1ea5u \u201c=\u201d x\u1ea3y ra khi: $\\left\\{ \\begin{aligned} & x+2=0 \\\\ & y-1=0 \\\\ & z-2=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x=-2 \\\\ & y=1 \\\\ & z=2 \\\\ \\end{aligned} \\right.$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 \u0111\u00e1p \u00e1n ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1$. <\/span><\/span> "}],"id_ques":788},{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai11/lv3/img\/13.jpg' \/><\/center> Cho $x=y+1$ th\u00ec $M=x^3-y^3-3xy$ c\u00f3 gi\u00e1 tr\u1ecb l\u00e0:","select":["A. $1$","B. $2$","C. $3$ ","D. $4$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1:<\/b> Thay $x=y+1$ v\u00e0o $M$ v\u00e0 bi\u1ebfn \u0111\u1ed5i, r\u00fat g\u1ecdn theo $y$. <br\/> <b> B\u01b0\u1edbc 2:<\/b> K\u1ebft lu\u1eadn gi\u00e1 tr\u1ecb c\u1ee7a $M$ t\u00ecm \u0111\u01b0\u1ee3c. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta thay $x=y+1$ v\u00e0o $M$:<br\/> $\\begin{align} & M=x^3-y^3-3xy \\\\ & =(y+1)^3-y^3-3(y+1)y \\\\ & =y^3+3y^2+3y+1-y^3-3y^2-3y \\\\ & =1 \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n A. <\/span><\/span> ","column":2}],"id_ques":789},{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":" K\u1ebft qu\u1ea3 ph\u00e2n t\u00edch \u0111a th\u1ee9c $x^4-x^2-6$ th\u00e0nh nh\u00e2n t\u1eed l\u00e0","select":["A. $(x^2+x+1)(x^3-x-6)$","B. $(x^2-x-1)(x^3+x+6)$","C. $(x^2-2)(x^2+3)$ ","D. $(x^2+2)(x^2-3)$"],"hint":" T\u00e1ch $-x^2 = 2x^2 - 3x^2$ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1:<\/b> T\u00e1ch $-x^2 = 2x^2 - 3x^2$ <br\/> <b> B\u01b0\u1edbc 2:<\/b> Ti\u1ebfp t\u1ee5c ph\u00e2n t\u00edch b\u1eb1ng c\u00e1ch nh\u00f3m v\u00e0 \u0111\u1eb7t nh\u00e2n t\u1eed chung. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> ${{x}^{4}}-{{x}^{2}}-6 \\\\ ={{x}^{4}}+2{{x}^{2}}-3{{x}^{2}}-6 \\\\ ={{x}^{2}}\\left( {{x}^{2}}+2 \\right)-3\\left( {{x}^{2}}+2 \\right) \\\\ =\\left( {{x}^{2}}+2 \\right)\\left( {{x}^{2}}-3 \\right)$ <br\/> <span class='basic_pink'> V\u1eady ch\u1ecdn \u0111\u00e1p \u00e1n D. <\/span><\/span> ","column":2}],"id_ques":790}],"id_ques":0}],"lesson":{"save":1,"level":3,"time":44}}