{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> Kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai7/lv2/img\/2.jpg' \/><\/center>V\u1edbi $x\\,\\,\\in\\mathbb{Z}$ th\u00ec $x^8+x^4+1$ l\u00e0 m\u1ed9t s\u1ed1 nguy\u00ean d\u01b0\u01a1ng. ","select":["\u0110\u00fang","Sai"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b> C\u00e1ch 1:<\/b> Ta \u0111\u00e1nh gi\u00e1 $x^8\\ge 0;x^4\\ge 0$ suy ra $x^8+x^4+1\\ge 1$. <br\/> <b> C\u00e1ch 2:<\/b><br\/> B\u01b0\u1edbc 1: Ph\u00e2n t\u00edch \u0111a th\u1ee9c theo c\u00e1ch t\u00e1ch: $x^8+x^4+1=x^8+2x^4+1-x^4$$=(x^8+2x^4+1)-x^4$.<br\/>B\u01b0\u1edbc 2: Ti\u1ebfp t\u1ee5c ph\u00e2n t\u00edch b\u1eb1ng c\u00e1ch d\u00f9ng H\u0110T. <br\/> <span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'> <b> C\u00e1ch 1:<\/b> Ta th\u1ea5y: $x^8\\ge 0$ v\u1edbi $\\forall x$; $x^4\\ge 0$ v\u1edbi $\\forall x$ suy ra $x^8+x^4+1\\ge 1 >0$.<br\/>Do \u0111\u00f3 \u0111a th\u1ee9c \u0111\u00e3 cho l\u00e0 m\u1ed9t s\u1ed1 nguy\u00ean d\u01b0\u01a1ng <br\/> <b> C\u00e1ch 2:<\/b><br\/>Ta c\u00f3:<br\/> $\\begin{align} &x^8+x^4+1 \\\\ & = x^8+2x^4+1-x^4 \\\\ & =(x^8+2x^4+1)-x^4 \\\\ & =(x^4+1)^2-(x^2)^2 \\\\ & =(x^4+1+x^2)(x^4+1-x^2) \\\\ \\end{align}$<br\/> Ta th\u1ea5y $x^4+x^2+1>0$, $x^4+1-x^2>0$ .<br\/> Do \u0111\u00f3 \u0111a th\u1ee9c \u0111\u00e3 cho l\u00e0 m\u1ed9t s\u1ed1 nguy\u00ean d\u01b0\u01a1ng <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang <\/span>","column":2}]}],"id_ques":601},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> Kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai7/lv2/img\/16.jpg' \/><\/center>V\u1edbi $x,y,z \\,\\,\\in\\mathbb{Z}$ th\u00ec $\\dfrac{27}{64}{{x}^{3}}{{y}^{6}}+\\dfrac{9}{8}{{x}^{2}}{{y}^{4}}{{z}^{2}}$$+x{{y}^{2}}{{z}^{4}}+\\dfrac{8}{27}{{z}^{6}}$ l\u00e0 l\u1eadp ph\u01b0\u01a1ng c\u1ee7a m\u1ed9t t\u1ed5ng. ","select":["\u0110\u00fang","Sai"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/>Ph\u00e2n t\u00edch \u0111a th\u1ee9c t\u1eebng b\u01b0\u1edbc theo l\u1eadp ph\u01b0\u01a1ng c\u1ee7a m\u1ed9t t\u1ed5ng. <br\/> <span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'> Ta c\u00f3:<br\/> $\\dfrac{27}{64}{{x}^{3}}{{y}^{6}}+\\dfrac{9}{8}{{x}^{2}}{{y}^{4}}{{z}^{2}}$$+x{{y}^{2}}{{z}^{4}}+\\dfrac{8}{27}{{z}^{6}} $<br\/>$ ={{\\left( \\dfrac{3}{4}x{{y}^{2}} \\right)}^{3}}+3.{{\\left( \\dfrac{3}{4}x{{y}^{2}} \\right)}^{2}}.\\left( \\dfrac{2}{3}{{z}^{2}} \\right)$$+3.\\left( \\dfrac{3}{4}x{{y}^{2}} \\right).{{\\left( \\dfrac{2}{3}{{z}^{2}} \\right)}^{2}}$$+{{\\left( \\dfrac{2}{3}{{z}^{2}} \\right)}^{3}} $<br\/>$ ={{\\left( \\dfrac{3}{4}x{{y}^{2}}+\\dfrac{2}{3}{{z}^{2}} \\right)}^{3}} $ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang <\/span>","column":2}]}],"id_ques":602},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai7/lv2/img\/16.jpg' \/><\/center> Ph\u00e2n t\u00edch \u0111a th\u1ee9c $x+2a(x-y)-y$ th\u00e0nh nh\u00e2n t\u1eed th\u00ec s\u1ebd c\u00f3 m\u1ed9t nh\u00e2n t\u1eed l\u00e0: ","select":["A. $1+2a$ ","B. $x+y$ ","C. $y-x$","D. $2a-2$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ta nh\u00f3m :<br\/>$\\text{x+2a}\\left( x-y \\right)-y $$=\\left( x-y \\right)+2a\\left( x-y \\right)$. <br\/><b> B\u01b0\u1edbc 2: <\/b>Ti\u1ebfp t\u1ee5c ph\u00e2n t\u00edch b\u1eb1ng c\u00e1ch \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'>Ta c\u00f3 :<br\/> $\\begin{align} & \\text{x+2a}\\left( x-y \\right)-y \\\\ & =\\left( x-y \\right)+2a\\left( x-y \\right) \\\\ & =\\left( x-y \\right)\\left( 1+2a \\right) \\\\ \\end{align}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A<\/span>","column":2}]}],"id_ques":603},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai7/lv2/img\/16.jpg' \/><\/center> Ph\u00e2n t\u00edch \u0111a th\u1ee9c $x^2-(a+b)x+ab$ th\u00e0nh nh\u00e2n t\u1eed th\u00ec s\u1ebd c\u00f3 m\u1ed9t nh\u00e2n t\u1eed l\u00e0: ","select":["A. $a+b$ ","B. $x-a$ ","C. $x+b$","D. $x+a$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ta nh\u00f3m :<br\/>${{x}^{2}}-(a+b)x+ab $$=\\left( {{x}^{2}}-bx \\right)-\\left( a\\,x-ab \\right)$. <br\/><b> B\u01b0\u1edbc 2: <\/b>Ti\u1ebfp t\u1ee5c ph\u00e2n t\u00edch b\u1eb1ng c\u00e1ch \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'>Ta c\u00f3 :<br\/> $\\begin{align} & {{x}^{2}}-(a+b)x+ab \\\\ & ={{x}^{2}}-a\\,x-bx+ab \\\\ & =\\left( {{x}^{2}}-bx \\right)-\\left( a\\,x-ab \\right) \\\\ & =x\\left( x-b \\right)-a\\left( x-b \\right) \\\\ & =\\left( x-b \\right)\\left( x-a \\right) \\\\ \\end{align}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B<\/span>","column":2}]}],"id_ques":604},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai7/lv2/img\/12.jpg' \/><\/center> Ph\u00e2n t\u00edch \u0111a th\u1ee9c $(a+b)^2-m^2+a+b-m$ th\u00e0nh nh\u00e2n t\u1eed th\u00ec s\u1ebd c\u00f3 m\u1ed9t nh\u00e2n t\u1eed l\u00e0: ","select":["A. $a+b-m$ ","B. $a+b+m$ ","C. $a+b$","D. $a-b-m$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ta nh\u00f3m :<br\/>${{(a+b)}^{2}}-{{m}^{2}}+a+b-m $$=\\left[ {{(a+b)}^{2}}-{{m}^{2}} \\right]+\\left( a+b-m \\right)$. <br\/><b> B\u01b0\u1edbc 2: <\/b>Ti\u1ebfp t\u1ee5c ph\u00e2n t\u00edch b\u1eb1ng c\u00e1ch d\u00f9ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c r\u1ed3i \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'>Ta c\u00f3 :<br\/> $ {{(a+b)}^{2}}-{{m}^{2}}+a+b-m $<br\/>$ =\\left[ {{(a+b)}^{2}}-{{m}^{2}} \\right]+\\left( a+b-m \\right) $<br\/>$ =\\left( a+b+m \\right)\\left( a+b-m \\right)$$+\\left( a+b-m \\right) $<br\/>$ =\\left( a+b-m \\right)\\left( a+b+m+1 \\right) $ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A<\/span>","column":2}]}],"id_ques":605},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai7/lv2/img\/12.jpg' \/><\/center> \u0110a th\u1ee9c $A=-7x^2+5xy+12y^2$ c\u00f3 gi\u00e1 tr\u1ecb b\u1eb1ng 0 th\u00ec m\u1ed1i quan h\u1ec7 gi\u1eefa x, y l\u00e0:","select":["A. $x=y$ ","B. $x-y=0$ ho\u1eb7c $12y+7x=0$ ","C. $x+y=0$ ho\u1eb7c $12y=7x$","D. $x+y=1$ ho\u1eb7c $12y=7x$"],"hint":" Ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch t\u00e1ch<br\/> T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 A = 0. N\u1ebfu a.b = 0 th\u00ec ho\u1eb7c a = 0 ho\u1eb7c b = 0.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed $-7x^2+5xy+12y^2$$=-7x^2-7xy+12xy+12y^2$. <br\/> <b> B\u01b0\u1edbc 2: <\/b>Ti\u1ebfp t\u1ee5c ph\u00e2n t\u00edch b\u1eb1ng c\u00e1ch \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/><b> B\u01b0\u1edbc 3:<\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 A = 0.<br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'>Ta c\u00f3 :<br\/> $\\begin{align} & A=-7x^2+5xy+12y^2 \\\\ & =-7x^2-7xy+12xy+12y^2 \\\\ & =(-7x^2-7xy)+(12xy+12y^2) \\\\ & =-7x(x+y)+12y(x+y) \\\\ & =(x+y)(-7x+12y) \\\\ \\end{align}$ <br\/>\u0110\u1ec3 A = 0 th\u00ec $\\left[ \\begin{align} & x+y=0 \\\\ & -7x+12y=0 \\\\ \\end{align} \\right.$<br\/> Do \u0111\u00f3 $x+y=0$ ho\u1eb7c $12y=7x$ th\u00ec A = 0<span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C<\/span>","column":2}]}],"id_ques":606},{"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/lop8/toan/daiso/bai7/lv2/img\/11.jpg' \/><\/center> \u0110a th\u1ee9c $A=3x^2(x+1)-5x(x+1)^2$$+4(x+1)$ c\u00f3 gi\u00e1 tr\u1ecb b\u1eb1ng $0$ th\u00ec: ","select":["A. $x = -1$ ","B. $x=-2$ ","C. $x=-3$ ","D. $x=-41$"],"hint":" Ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed<br\/>T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 $A = 0$. N\u1ebfu $a.b = 0 $ th\u00ec ho\u1eb7c $a = 0$ ho\u1eb7c $b = 0$.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed <br\/>$3x^2(x+1)-5x(x+1)^2+4(x+1)$$=(x+1)(3x^2-5x^2-5x+4)$$=(x+1)(-2x^2-5x+4)$. <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 $A = 0$. <br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'>Ta c\u00f3 :<br\/> $ A=3x^2(x+1)-5x(x+1)^2$$+4(x+1) $<br\/>$ =(x+1)(3x^2-5x^2-5x+4) $<br\/>$ =(x+1)(-2x^2-5x+4) $ <br\/>\u0110\u1ec3 $A = 0$ th\u00ec $\\left[ \\begin{align} & x+1=0 \\\\ & -2x^2-5x+4=0 \\\\ \\end{align} \\right.$ <br\/> $\\Rightarrow$ x = -1<br\/> <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A<\/span>","column":2}]}],"id_ques":607},{"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/lop8/toan/daiso/bai7/lv2/img\/11.jpg' \/><\/center> \u0110a th\u1ee9c $A=x^2-7xy+10y^2$ c\u00f3 gi\u00e1 tr\u1ecb b\u1eb1ng 0 th\u00ec m\u1ed1i quan h\u1ec7 gi\u1eefa $x$ v\u00e0 $y$ l\u00e0:","select":["A. $x = 5y$ ho\u1eb7c $x =2y$ ","B. $x=y$ ","C. $x=2y$ ho\u1eb7c $x=4y$","D. $x=-2y$ ho\u1eb7c $x=-3y$"],"hint":" Ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n b\u1eb1ng c\u00e1ch t\u00e1ch r\u1ed3i \u0111\u1eb7t nh\u00e2n t\u1eed chung. <br\/> T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 $A = 0$. N\u1ebfu $a.b = 0 $ th\u00ec ho\u1eb7c $a = 0$ ho\u1eb7c $b = 0$.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed $x^2-7xy+10y^2$$=x^2-2xy-5xy+10y^2$$=(x^2-2xy)-(5xy-10y^2)$. <br\/> <b> B\u01b0\u1edbc 2: <\/b>Ti\u1ebfp t\u1ee5c ph\u00e2n t\u00edch b\u1eb1ng c\u00e1ch \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/><b> B\u01b0\u1edbc 3:<\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 $A = 0$. N\u1ebfu $a.b = 0 $ th\u00ec ho\u1eb7c $a = 0$ ho\u1eb7c $b = 0$.<br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'>Ta c\u00f3 :<br\/> $\\begin{align} & A=x^2-7xy+10y^2 \\\\ & =x^2-2xy-5xy+10y^2 \\\\ & =(x^2-2xy)-(5xy-10y^2) \\\\ & =x(x-2y)-5y(x-2y) \\\\ & =(x-2y)(x-5y) \\\\\\end{align}$ <br\/>\u0110\u1ec3 $A = 0$ th\u00ec $\\left[ \\begin{align} & x-2y=0 \\\\ & x-5y=0 \\\\ \\end{align} \\right.$<br\/> Do \u0111\u00f3 $x = 2y$ ho\u1eb7c $x = 5y$ th\u00ec $A = 0$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A<\/span>","column":2}]}],"id_ques":608},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["0"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai7/lv2/img\/10.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a \u0111a th\u1ee9c $x^5-5x^3+4x$ v\u1edbi $x = 2$ l\u00e0 _input_","hint":" <b> C\u00e1ch 1:<\/b> Thay $x = 2$ v\u00e0o \u0111a th\u1ee9c r\u1ed3i t\u00ednh.<br\/> <b> C\u00e1ch 2:<\/b> Ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed <br\/> Thay $x = 2$ v\u00e0o \u0111a th\u1ee9c \u0111\u1ec3 t\u00ednh gi\u00e1 tr\u1ecb.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> C\u00e1ch 1:<\/b> Thay $x = 2$ v\u00e0o \u0111a th\u1ee9c r\u1ed3i t\u00ednh.<br\/> <b> C\u00e1ch 2:<\/b> <br\/>B\u01b0\u1edbc 1: Ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed $x^5-5x^3+4x$$=x(x^4-5x^2+4)$$=x(x^4-x^2-4x^2+4)$.<br\/> B\u01b0\u1edbc 2: Ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/> B\u01b0\u1edbc 3: Thay $x = 2$ v\u00e0o \u0111a th\u1ee9c \u0111\u1ec3 t\u00ednh gi\u00e1 tr\u1ecb.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/><b> C\u00e1ch 1:<\/b> Thay $x = 2$ v\u00e0o \u0111a th\u1ee9c ta \u0111\u01b0\u1ee3c:<br\/> $x^5-5x^3+4x=2^5-5.2^3+4.2$$=32-40+8=0$.<br\/> <b> C\u00e1ch 2:<\/b><br\/>Ta c\u00f3:<br\/> $\\begin{align} & {{x}^{5}}-5{{x}^{3}}+4x \\\\ & =x({{x}^{4}}-5{{x}^{2}}+4) \\\\ & =x({{x}^{4}}-{{x}^{2}}-4{{x}^{2}}+4) \\\\ & =x\\left[ \\left( {{x}^{4}}-{{x}^{2}} \\right)-\\left( 4{{x}^{2}}-4 \\right) \\right] \\\\ & =x\\left[ {{x}^{2}}\\left( {{x}^{2}}-1 \\right)-4\\left( {{x}^{2}}-1 \\right) \\right] \\\\ & =x\\left( {{x}^{2}}-1 \\right)\\left( {{x}^{2}}-4 \\right) \\\\ & =x\\left( x+1 \\right)\\left( x-1 \\right)\\left( x+2 \\right)\\left( x-2 \\right) \\\\ \\end{align}$ <br\/>Thay x = 2 v\u00e0o \u0111a th\u1ee9c \u0111\u00e3 r\u00fat g\u1ecdn: <br\/> $x\\left( x+1 \\right)\\left( x-1 \\right)\\left( x+2 \\right)\\left( x-2 \\right)$$=2.\\left( 2+1 \\right)\\left( 2-1 \\right)\\left( 2+2 \\right)\\left( 2-2 \\right)=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":609},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["-672"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai7/lv2/img\/9.jpg' \/><\/center> Ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed r\u1ed3i th\u1ef1c hi\u1ec7n t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a \u0111a th\u1ee9c $x^3-7x-6$ <br\/>v\u1edbi $x = -9$<br\/>Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c l\u00e0: _input_<\/span>","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed $x^3-7x-6=x^3-x-6x-6$$=(x^3-x)-(6x+6)$.<br\/> <b>B\u01b0\u1edbc 2:<\/b> Ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/> <b> B\u01b0\u1edbc 3:<\/b> Thay $x = -9$ v\u00e0o \u0111a th\u1ee9c \u0111\u1ec3 t\u00ednh gi\u00e1 tr\u1ecb.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Ta c\u00f3:<br\/> $\\begin{align} & x^3-7x-6 \\\\ & =x^3-x-6x-6 \\\\ & =(x^3-x)-(6x+6) \\\\ & =x(x^2-1)-6(x+1) \\\\ & =x(x-1)(x+1)-6(x+1) \\\\ & =(x+1)(x^2-x-6) \\\\ & =(x+1)(x^2+2x-3x-6) \\\\ & =(x+1)[(x^2+2x)-(3x+6)] \\\\ & =(x+1)[x(x+2)-3(x+2)] \\\\ & =(x+1)(x+2)(x-3) \\\\ \\end{align}$ <br\/>Thay $x = -9$ v\u00e0o \u0111a th\u1ee9c, ta \u0111\u01b0\u1ee3c: <br\/>$(x+1)(x+2)(x-3)$$=(-9+1)(-9+2)(-9-3)$$=-8.(-7).(-12)=-672$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-672$ <\/span><\/span> "}]}],"id_ques":610},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["2500"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai7/lv2/img\/9.jpg' \/><\/center>Gi\u00e1 tr\u1ecb c\u1ee7a \u0111a th\u1ee9c<br\/> ${{x}^{2}}+\\dfrac{1}{2}x+\\dfrac{1}{16}$ v\u1edbi $x = 49,75$ l\u00e0: _input_","hint":" Ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed<br\/> Thay $x = 49,75$ v\u00e0o \u0111a th\u1ee9c \u0111\u1ec3 t\u00ednh gi\u00e1 tr\u1ecb.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed ${{x}^{2}}+\\dfrac{1}{2}x+\\dfrac{1}{16}$$={{\\left( x+\\dfrac{1}{4} \\right)}^{2}}$.<br\/> <b>B\u01b0\u1edbc 2:<\/b> Ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/> <b> B\u01b0\u1edbc 3:<\/b> Thay $x = 49,75$ v\u00e0o \u0111a th\u1ee9c \u0111\u1ec3 t\u00ednh gi\u00e1 tr\u1ecb.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Ta c\u00f3:<br\/> $\\begin{align} & {{x}^{2}}+\\dfrac{1}{2}x+\\dfrac{1}{16} \\\\ & ={{x}^{2}}+2.x.\\dfrac{1}{4}+{{\\left( \\dfrac{1}{4} \\right)}^{2}} \\\\ & ={{\\left( x+\\dfrac{1}{4} \\right)}^{2}} \\\\ \\end{align}$ <br\/>Thay $x = 49,75$ v\u00e0o \u0111a th\u1ee9c ta \u0111\u01b0\u1ee3c: <br\/>${{\\left( x+\\dfrac{1}{4} \\right)}^{2}}={{\\left( 49,75+0,25 \\right)}^{2}}$$={{50}^{2}}=2500$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2500$ <\/span><\/span> "}]}],"id_ques":611},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank_random","correct":[[["2"],["3"],["1"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai7/lv2/img\/8.jpg' \/><\/center> Bi\u1ebft $a^3-6a^2+11a-6=0$, khi \u0111\u00f3 $\\left[ \\begin{align} & a= \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\\\ & a=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & a=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{align} \\right.$ ","hint":"Ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch t\u00e1ch r\u1ed3i nh\u00f3m<br\/> N\u1ebfu $a.b = 0$ th\u00ec $ a = 0$ ho\u1eb7c $b = 0$. ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/> <b>B\u01b0\u1edbc 1:<\/b> Ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch t\u00e1ch r\u1ed3i nh\u00f3m: <br\/>$a^3-6a^2+11a-6$$=a^3-a^2-5a^2+5a+6a-6$$=(a^3-a^2)-(5a^2-5a)+(6a-6)$.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Ph\u00e2n t\u00edch $(a^3-a^2)-(5a^2-5a)+(6a-6)$ th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ecm $a$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta c\u00f3:<br\/> $ a^3-6a^2+11a-6=0 $<br\/>$\\Leftrightarrow$ $ a^3-a^2-5a^2+5a$$+6a-6=0 $<br\/>$\\Leftrightarrow$ $(a^3-a^2)-(5a^2-5a)$$+(6a-6) $<br\/>$\\Leftrightarrow$ $ a^2(a-1)-5a(a-1)$$+6(a-1)=0 $<br\/>$\\Leftrightarrow$ $ (a-1)(a^2-5a+6)=0 $<br\/>$\\Leftrightarrow$ $ (a-1)(a^2-2a-3a+6)=0 $<br\/>$\\Leftrightarrow$ $ (a-1)[a(a-2)-3(a-2)]=0 $<br\/>$\\Leftrightarrow$ $ (a-1)(a-2)(a-3)=0 $<br\/>$ \\Rightarrow \\left[ \\begin{aligned} & a-1=0 \\\\ & a-2=0 \\\\ & a-3=0 \\\\ \\end{aligned} \\right.$$\\Leftrightarrow \\left[ \\begin{aligned} & a=1 \\\\ & a=2 \\\\ & a=3 \\\\ \\end{aligned} \\right. $ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1, 2 $ v\u00e0 $3$ <\/span><\/span> "}]}],"id_ques":612},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["0"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai7/lv2/img\/5.jpg' \/><\/center> Bi\u1ebft $x^3+3x^2+3x=0$, khi \u0111\u00f3 x l\u00e0 _input_ ","hint":"Ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed<br\/> N\u1ebfu $a.b = 0$ th\u00ec $a = 0$ ho\u1eb7c $b = 0$. ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b>B\u01b0\u1edbc 1:<\/b> Ph\u00e2n t\u00edch v\u1ebf tr\u00e1i : $x^3+3x^2+3x$$=x(x^2+3x+3)$.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Ch\u1ee9ng minh $x^2+3x+3>0$.<br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ecm $x$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta c\u00f3:<br\/> $\\begin{align} & {{x}^{3}}+3{{x}^{2}}+3x=0 \\\\ & \\Leftrightarrow x\\left( {{x}^{2}}+3x+3 \\right)=0 \\\\ & \\text{Do} \\,\\,\\,{{x}^{2}}+3x+3\\,\\,>\\,\\,0 \\\\ & \\Rightarrow x=0 \\\\ \\end{align}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0$ <\/span><\/span> "}]}],"id_ques":613},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\left[\\begin{array}{l}{x=\\dfrac{-7}{2}} \\\\ {x = \\dfrac{-1}{4}}\\end{array}\\right.$","B. $\\left[\\begin{array}{l}{x=\\dfrac{7}{2}} \\\\ {x = \\dfrac{-1}{2}}\\end{array}\\right.$","C. $\\left[\\begin{array}{l}{x=\\dfrac{5}{2}} \\\\ {x = \\dfrac{-1}{2}}\\end{array}\\right.$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai7/lv2/img\/5.jpg' \/><\/center> Bi\u1ebft $8x^2+30x+7=0$, gi\u00e1 tr\u1ecb $x =?$","hint":"Ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch t\u00e1ch r\u1ed3i nh\u00f3m<br\/> N\u1ebfu $a.b = 0$ th\u00ec $a = 0$ ho\u1eb7c $b = 0$. ","explain":"<span class='basic_left'><span class='basic_green'>Ghi nh\u1edb<\/span><br\/> Ph\u00e2n t\u00edch \u0111a th\u1ee9c $ax^2+bx+c$ th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p t\u00e1ch.<br\/> Ta t\u00e1ch: $ax^2+bx+c$$=ax^2+mx+nx+c$.<br\/> Ta t\u00ecm m, n t\u1eeb: $\\left\\{ \\begin{align} & m+n=b \\\\ & m.n=a.c \\\\ \\end{align} \\right.$<br\/><span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b>B\u01b0\u1edbc 1:<\/b> Ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch t\u00e1ch r\u1ed3i nh\u00f3m: <br\/>$8x^2+30x+7$$=(8x^2+28x)+(2x+7)$.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Ph\u00e2n t\u00edch $(8x^2+28x)+(2x+7)$ th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ecm x <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta c\u00f3:<br\/> $\\begin{aligned} & 8x^2+30x+7=0 \\\\ & \\Leftrightarrow (8x^2+28x)+(2x+7)=0 \\\\ & \\Leftrightarrow 4x(2x+7)+(2x+7)=0 \\\\ & \\Leftrightarrow (2x+7)(4x+1)=0 \\\\ & \\Rightarrow \\left[ \\begin{aligned} & 2x+7=0 \\\\ & 4x+1=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & x=\\dfrac{-7}{2} \\\\ & x=\\dfrac{-1}{4} \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<\/span> "}]}],"id_ques":614},{"time":24,"part":[{"title":"\u0110i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng trong k\u1ebft qu\u1ea3 ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed","title_trans":"","temp":"fill_the_blank_random","correct":[[["x-2y"],["x-y"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai7/lv2/img\/4.jpg' \/><\/center> $x^2-3xy+2y^2$$=(\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{})\\times(\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{})$","hint":"Ta ph\u00e2n t\u00edch b\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p t\u00e1ch r\u1ed3i nh\u00f3m <br\/>Ph\u00e2n t\u00edch ti\u1ebfp b\u1eb1ng c\u00e1ch \u0111\u1eb7t nh\u00e2n t\u1eed chung.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b> B\u01b0\u1edbc 1:<\/b> Ta ph\u00e2n t\u00edch b\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p t\u00e1ch r\u1ed3i nh\u00f3m: $x^2-2xy-xy+2y^2=(x^2-2xy)-(xy-2y^2)$.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Ph\u00e2n t\u00edch ti\u1ebfp theo ph\u01b0\u01a1ng ph\u00e1p \u0111\u1eb7t nh\u00e2n t\u1eed chung. <br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'>Ta c\u00f3:<br\/> $\\begin{align} & x^2-3xy+2y^2 \\\\ & =x^2-2xy-xy+2y^2 \\\\ & =(x^2-2xy)-(xy-2y^2) \\\\ & =x\\left( x-2y \\right)-y\\left( x-2y \\right) \\\\ & =\\left( x-2y \\right)\\left( x-y \\right) \\\\ \\end{align}$<span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0 $x - y$ v\u00e0 $x - 2y$<\/span>"}]}],"id_ques":615},{"time":24,"part":[{"title":"\u0110i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng trong k\u1ebft qu\u1ea3 ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed","title_trans":"","temp":"fill_the_blank_random","correct":[[["5x+2"],["2+5x"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai7/lv2/img\/4.jpg' \/><\/center> $5x^2-18x-8$$=(x-4)(\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{})$","hint":"Ta ph\u00e2n t\u00edch b\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p t\u00e1ch r\u1ed3i nh\u00f3m<br\/>Ph\u00e2n t\u00edch ti\u1ebfp b\u1eb1ng c\u00e1ch \u0111\u1eb7t nh\u00e2n t\u1eed chung.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b> B\u01b0\u1edbc 1:<\/b> Ta ph\u00e2n t\u00edch b\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p t\u00e1ch r\u1ed3i nh\u00f3m: $5x^2-20x+2x-8$$=(5x^2-20x)+(2x-8)$.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Ph\u00e2n t\u00edch ti\u1ebfp theo ph\u01b0\u01a1ng ph\u00e1p \u0111\u1eb7t nh\u00e2n t\u1eed chung. <br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'>Ta c\u00f3:<br\/> $\\begin{align} & 5x^2-18x-8 \\\\ & =5x^2-20x+2x-8 \\\\ & =(5x^2-20x)+(2x-8) \\\\ & =5x\\left( x-4 \\right)+2\\left( x-4 \\right) \\\\ & =\\left( x-4 \\right)\\left( 5x+2 \\right) \\\\ \\end{align}$<span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0 $5x + 2$<\/span>"}]}],"id_ques":616},{"time":24,"part":[{"title":"\u0110i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng trong k\u1ebft qu\u1ea3 ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed","title_trans":"","temp":"fill_the_blank_random","correct":[[["x+2"],["2+x"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai7/lv2/img\/3.jpg' \/><\/center> $4x^2+5x-6$$=(4x-3)(\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{})$","hint":"Ta ph\u00e2n t\u00edch b\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p t\u00e1ch r\u1ed3i nh\u00f3m <br\/>Ph\u00e2n t\u00edch ti\u1ebfp b\u1eb1ng c\u00e1ch \u0111\u1eb7t nh\u00e2n t\u1eed chung.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b> B\u01b0\u1edbc 1:<\/b> Ta ph\u00e2n t\u00edch b\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p t\u00e1ch r\u1ed3i nh\u00f3m: $4x^2+8x-3x-6$$=(4x^2+8x)-(3x+6)$.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Ph\u00e2n t\u00edch ti\u1ebfp theo ph\u01b0\u01a1ng ph\u00e1p \u0111\u1eb7t nh\u00e2n t\u1eed chung. <br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'>Ta c\u00f3:<br\/> $\\begin{align} & 4x^2+5x-6 \\\\ & =4x^2+8x-3x-6 \\\\ & =(4x^2+8x)-(3x+6) \\\\ & =4x\\left( x+2 \\right)-3\\left( x+2 \\right) \\\\ & =\\left( x+2 \\right)\\left( 4x-3 \\right) \\\\ \\end{align}$<span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0 $x + 2$<\/span>"}]}],"id_ques":617},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai7/lv2/img\/3.jpg' \/><\/center> Ph\u00e2n t\u00edch \u0111a th\u1ee9c $4x^4+81$ th\u00e0nh nh\u00e2n t\u1eed, ta \u0111\u01b0\u1ee3c:","select":["A. $(2x^2-6x+9)(2x^2+6x+9)$ ","B. $(2x^2+6x-9)(2x^2-6x+9)$ ","C. $(2x^2-9)(2x^2+9)$","D. $(2x^2+9)(2x^2+6x+9)$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/>Ta th\u00eam, b\u1edbt $36x^2$ \u0111\u1ec3 \u0111\u01b0a \u0111a th\u1ee9c v\u1ec1 hi\u1ec7u hai b\u00ecnh ph\u01b0\u01a1ng: $4x^4+36x^2+81-36x^2$$=(2x^2+9)^2-36x^2$. <br\/>Ph\u00e2n t\u00edch ti\u1ebfp b\u1eb1ng c\u00e1ch d\u00f9ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c .<br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'>Ta c\u00f3:<br\/> $\\begin{align} & 4x^4+81 \\\\ & = 4x^4+36x^2+81-36x^2 \\\\ & =(x^2)^2+2.2x^2.9+9^2-36x^2 \\\\ & =(2x^2+9)^2-(6x)^2 \\\\ & =(2x^2+9+6x)(2x^2+9-6x) \\\\ \\end{align}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A<\/span>","column":2}]}],"id_ques":618},{"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/lop8/toan/daiso/bai7/lv2/img\/2.jpg' \/><\/center> Ph\u00e2n t\u00edch \u0111a th\u1ee9c $x^2(x^2+4)-x^2+4$ th\u00e0nh nh\u00e2n t\u1eed, ta \u0111\u01b0\u1ee3c:","select":["A. $(x^2+4-x)(x^2+4+x)$ ","B. $(x^2+4-x)^2$ ","C. $(x^2-x+2)(x^2+x+2)$","D. $(x^2+x+2)^2$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/>Ta nh\u00e2n ra v\u00e0 nh\u00f3m: $x^2(x^2+4)-x^2+4$$=(x^4+4x^2+4)-x^2$ <br\/>Ph\u00e2n t\u00edch ti\u1ebfp b\u1eb1ng c\u00e1ch d\u00f9ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c .<br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'>Ta c\u00f3:<br\/> $\\begin{align} & x^2(x^2+4)-x^2+4 \\\\ & = x^4+4x^2-x^2+4 \\\\ & =x^4+4x^2+4-x^2 \\\\ & =(x^4+4x^2+4)-x^2 \\\\ & =(x^2+2)^2-x^2 \\\\ & =(x^2+2+x)(x^2+2-x) \\\\ \\end{align}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C<\/span>","column":2}]}],"id_ques":619},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai7/lv2/img\/2.jpg' \/><\/center> Ph\u00e2n t\u00edch \u0111a th\u1ee9c $2x^2-7xy+5y^2$ th\u00e0nh nh\u00e2n t\u1eed, ta \u0111\u01b0\u1ee3c:","select":["A. $(x+y)(2x+5y)$ ","B. $(x-y)(x-5y)$ ","C. $(x+y)(2x-5y)$","D. $(x-y)(2x-5y)$"],"hint":" Ta ph\u00e2n t\u00edch b\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p t\u00e1ch r\u1ed3i nh\u00f3m","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/>Ta ph\u00e2n t\u00edch b\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p t\u00e1ch r\u1ed3i nh\u00f3m: <br\/>$2x^2-7xy+5y^2$$=2x^2-2xy-5xy+5y^2$$=(2x^2-2xy)-(5xy-5y^2)$ <br\/>Ph\u00e2n t\u00edch ti\u1ebfp b\u1eb1ng c\u00e1ch \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'>Ta c\u00f3:<br\/> $\\begin{align} & 2{{x}^{2}}-7xy+5{{y}^{2}} \\\\ & =2{{x}^{2}}-2xy-5xy+5{{y}^{2}} \\\\ & =(2{{x}^{2}}-2xy)-(5xy-5{{y}^{2}}) \\\\ & =2x\\left( x-y \\right)-5y\\left( x-y \\right) \\\\ & =\\left( x-y \\right)\\left( 2x-5y \\right) \\\\ \\end{align}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D<\/span>","column":2}]}],"id_ques":620}],"lesson":{"save":0,"level":2}}