{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","temp":"true_false","correct":[["t","t","f"]],"list":[{"point":10,"image":"https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai6/lv3/img\/1.png","col_name":["Kh\u1eb3ng \u0111\u1ecbnh","\u0110\u00fang","Sai"],"arr_ques":["Gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c $(x+1)^3-(5+3x+3x^2+x^3)$ kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o $x$ ","V\u1edbi $x\\in\\mathbb{N}$ th\u00ec bi\u1ec3u th\u1ee9c $x^3+3x^2+2x$ l\u00e0 t\u00edch c\u1ee7a ba s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp","N\u1ebfu $x = 1; y = 2$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $8x^3-12x^2y+6xy^2-y^3$ b\u1eb1ng $1$"],"hint":"<br\/>G\u1ee3i \u00fd 1: Khai tri\u1ec3n v\u00e0 r\u00fat g\u1ecdn $(x+1)^3-(5+3x+3x^2+x^3)$, xem k\u1ebft qu\u1ea3 c\u00f3 ch\u1ee9a $x$ hay kh\u00f4ng.<br\/> G\u1ee3i \u00fd 2: Ph\u00e2n t\u00edch $x^3+3x^2+2x$ th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng c\u00e1ch \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/> G\u1ee3i \u00fd 3: R\u00fat g\u1ecdn $8x^3-12x^2y+6xy^2-y^3$ v\u00e0 thay $x = 1; y = 2$ v\u00e0o \u0111\u1ec3 t\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c. ","explain":["<span class='basic_left'> <b>\u0110\u00fang v\u00ec <\/b>: <br\/> $(x+1)^3-(5+3x+3x^2+x^3)=x^3+3x^2+3x+1-5-3x-3x^2-x^3=-4$<br\/> gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o $x$<br\/><\/span>","<span class='basic_left'><b>\u0110\u00fang v\u00ec:<\/b> $x^3+3x^2+2x=x(x^2+3x+2)$ <br\/>$=x(x^2+x+2x+2)=x[x(x+1)+2(x+1)]$<br\/>$=x(x+1)(x+2)$ <br\/> l\u00e0 t\u00edch c\u1ee7a ba s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp v\u1edbi $x\\in\\mathbb{N}$<br\/><\/span> ","<span class='basic_left'><b>Sai v\u00ec:<\/b> $8x^3-12x^2y+6xy^2-y^3=(2x-y)^3$<br\/> Thay $x = 1; y = 2$ v\u00e0o: $(2x-y)^3=(2.1-2)^3=0$ <br\/><\/span> "]}]}],"id_ques":571},{"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":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai6/lv3/img\/16.jpg' \/><\/center> $(5x+2)^2-4$ chia h\u1ebft cho $5$ v\u1edbi m\u1ecdi $x\\in \\mathbb{Z}$ ","select":["\u0110\u00fang","Sai"],"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 $(5x+2)^2-4$ th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c. <br\/> <b> B\u01b0\u1edbc 2:<\/b> K\u1ebft qu\u1ea3 ph\u00e2n t\u00edch c\u00f3 ch\u1ee9a nh\u00e2n t\u1eed chia h\u1ebft cho 5 th\u00ec $(5x+2)^2-4$ chia h\u1ebft cho $5$ v\u00e0 ng\u01b0\u1ee3c l\u1ea1i.<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'> Ta c\u00f3:<br\/> $\\begin{align} & {{(5x+2)}^{2}}-4 \\\\& =\\left( 5x+2+2 \\right)\\left( 5x+2-2 \\right)\\\\& =\\left( 5x+4 \\right)5x \\\\ & \\text{Do}\\,\\,5x\\,\\,\\,\\vdots \\,\\,\\,5\\,\\,\\Rightarrow \\,\\,\\left[ {{(5x+2)}^{2}}-4 \\right]\\,\\,\\,\\,\\vdots \\,\\,\\,\\,5 \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang .<\/span>","column":2}]}],"id_ques":572},{"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":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai6/lv3/img\/13.jpg' \/><\/center> Ph\u00e2n t\u00edch \u0111a th\u1ee9c $(a+b)^3+(c-a)^3-(b+c)^3$ th\u00e0nh nh\u00e2n t\u1eed r\u1ed3i t\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c t\u1ea1i $a = b = c = 2016$. <br\/> \u0110\u00e1p \u00e1n 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> Nh\u00f3m ${{(a+b)}^{3}}+{{(c-a)}^{3}}-{{(b+c)}^{3}} $$=\\left[ {{(a+b)}^{3}}+{{(c-a)}^{3}} \\right]-{{(b+c)}^{3}}$.<br\/> <b>B\u01b0\u1edbc 2:<\/b> Ti\u1ebfp t\u1ee5c ph\u00e2n t\u00edch th\u00e0nh nh\u00e2n t\u1eed theo h\u1eb1ng \u0111\u1eb3ng th\u1ee9c v\u00e0 \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/> <b> B\u01b0\u1edbc 3:<\/b> Thay $a = b = c = 2016$ v\u00e0o bi\u1ec3u 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} & {{(a+b)}^{3}}+{{(c-a)}^{3}}-{{(b+c)}^{3}} \\\\ & =\\left[ {{(a+b)}^{3}}+{{(c-a)}^{3}} \\right]-{{(b+c)}^{3}} \\\\ & =\\left[ \\left( a+b \\right)+\\left( c-a \\right) \\right]\\left[ {{\\left( a+b \\right)}^{2}}-\\left( a+b \\right)\\left( c-a \\right)+{{\\left( c-a \\right)}^{2}} \\right]-{{\\left( b+c \\right)}^{3}} \\\\ & =\\left( b+c \\right)\\left[ {{a}^{2}}+2ab+{{b}^{2}}-ac+{{a}^{2}}-bc+ab+{{c}^{2}}-2ac+{{a}^{2}} \\right]-{{\\left( b+c \\right)}^{3}} \\\\ & =\\left( b+c \\right)\\left( 3{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+3ab-3ac-bc \\right)-{{\\left( b+c \\right)}^{3}} \\\\ & =\\left( b+c \\right)\\left[ 3{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+3ab-3ac-bc-{{\\left( b+c \\right)}^{2}} \\right] \\\\ & =\\left( b+c \\right)\\left( 3{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+3ab-3ac-bc-{{b}^{2}}-2bc-{{c}^{2}} \\right) \\\\ & =\\left( b+c \\right)\\left( 3{{a}^{2}}+3ab-3ac-3bc \\right) \\\\ & =\\left( b+c \\right)3\\left( {{a}^{2}}+ab-ac-bc \\right) \\\\ & =3\\left( b+c \\right)\\left[ \\left( {{a}^{2}}-ac \\right)+\\left( ab-bc \\right) \\right] \\\\ & =3\\left( b+c \\right)\\left[ a\\left( a-c \\right)+b\\left( a-c \\right) \\right] \\\\ & =3\\left( b+c \\right)\\left( a+b \\right)\\left( a-c \\right) \\\\ \\end{align}$ <br\/>Thay $a = b = c = 2016$ v\u00e0o ta \u0111\u01b0\u1ee3c:<br\/> $3\\left( b+c \\right)\\left( a+b \\right)\\left( a-c \\right)=3\\left( 2016+2016 \\right)\\left( 2016+2016 \\right)\\left( 2016-2016 \\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":573},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["68"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai6/lv3/img\/12.jpg' \/><\/center> Ph\u00e2n t\u00edch \u0111a th\u1ee9c $81y^2-(y^2+6y)^2$ th\u00e0nh nh\u00e2n t\u1eed r\u1ed3i t\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c t\u1ea1i $y = 2$. <br\/> \u0110\u00e1p \u00e1n 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> Khai tri\u1ec3n $81y^2-(y^2+6y)^2$ theo h\u1eb1ng \u0111\u1eb3ng th\u1ee9c hi\u1ec7u hai b\u00ecnh ph\u01b0\u01a1ng.<br\/> <b>B\u01b0\u1edbc 2:<\/b> Ti\u1ebfp t\u1ee5c ph\u00e2n t\u00edch th\u00e0nh nh\u00e2n t\u1eed.<br\/> <b> B\u01b0\u1edbc 3:<\/b> Thay $y = 2$ v\u00e0o bi\u1ec3u 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} & 81{{y}^{2}}-{{({{y}^{2}}+6y)}^{2}} \\\\ & ={{\\left( 9y \\right)}^{2}}-{{\\left( {{y}^{2}}+6y \\right)}^{2}} \\\\ & =\\left( 9y-{{y}^{2}}-6y \\right)\\left( 9y+{{y}^{2}}+6y \\right) \\\\ & =\\left( 3y-{{y}^{2}} \\right)\\left( {{y}^{2}}+15y \\right) \\\\ & =y\\left( 3-y \\right)y\\left( y+15 \\right) \\\\ & ={{y}^{2}}\\left( 3-y \\right)\\left( y+15 \\right) \\\\ \\end{align}$ <br\/>Thay $y = 2$ v\u00e0o bi\u1ec3u th\u1ee9c sau khi r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> ${{y}^{2}}\\left( 3-y \\right)\\left( y+15 \\right)={{2}^{2}}\\left( 3-2 \\right)\\left( 2+15 \\right)=4.17=68$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $68$. <\/span><\/span> "}]}],"id_ques":574},{"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":[[["y"],["3"],["x"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai6/lv3/img\/10.jpg' \/><\/center> $(x^2+y^2-5)^2-4(xy-2)^2$<br\/>$=(\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}+\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}+\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{})(x+y-3)(x-y+1)(x-y-1)$","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b> B\u01b0\u1edbc 1:<\/b> $(x^2+y^2-5)^2-4(xy-2)^2=(x^2+y^2-5)^2-[2(xy-2)]^2$, khai tri\u1ec3n theo h\u1eb1ng \u0111\u1eb3ng th\u1ee9c th\u1ee9 ba.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Ph\u00e2n t\u00edch ti\u1ebfp theo ph\u01b0\u01a1ng ph\u00e1p d\u00f9ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/> $\\begin{align} & (x^2 + y^2 - 5)^2 - 4(xy - 2)^2 \\\\ &= (x^2 + y^2 - 5)^2 - [2(xy - 2)]^2 \\\\ &= [x^2 + y^2 - 5 - 2(xy - 2)].[x^2 + y^2 - 5 + 2(xy - 2)] \\\\ &= [x^2 + y^2 - 2xy - 1].[x^2 + y^2 + 2xy - 9] \\\\ &= [(x - y)^2 - 1] . [(x + y)^2 - 9] \\\\ &= (x - y - 1)(x - y + 1)(x + y - 3)(x + y + 3) \\end{align}$<span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0 $x, y$ v\u00e0 $3$.<\/span>"}]}],"id_ques":575},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank_random","correct":[[["2"],["1"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai6/lv3/img\/9.jpg' \/><\/center> Bi\u1ebft $x^3-x^2=4x^2-8x+4$, khi \u0111\u00f3 $\\left[ \\begin{align} & x= \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\\\ & x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{align} \\right.$ ","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 $x^3-x^2$ th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng \u0111\u1eb7t $x^2$ l\u00e0m nh\u00e2n t\u1eed chung.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Ti\u1ebfp t\u1ee5c ph\u00e2n t\u00edch $4x^4-8x+4=4(x-2)^2$ .<br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ecm $x$ N\u1ebfu $a.b = 0$ th\u00ec $a = 0$ ho\u1eb7c $b = 0$. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3:<br\/> $\\begin{aligned} & {{x}^{3}}-{{x}^{2}}=4{{x}^{2}}-8x+4 \\\\ &\\Leftrightarrow {{x}^{3}}-{{x}^{2}}=4\\left( {{x}^{2}}-2x+1 \\right) \\\\ &\\Leftrightarrow {{x}^{2}}\\left( x-1 \\right)=4{{\\left( x-1 \\right)}^{2}} \\\\ &\\Leftrightarrow {{x}^{2}}\\left( x-1 \\right)-4{{\\left( x-1 \\right)}^{2}}=0 \\\\ &\\Leftrightarrow \\left( x-1 \\right)\\left[ {{x}^{2}}-4\\left( x-1 \\right) \\right]=0 \\\\ &\\Leftrightarrow \\left( x-1 \\right)\\left( {{x}^{2}}-4x+4 \\right)=0 \\\\ &\\Leftrightarrow \\left( x-1 \\right){{\\left( x-2 \\right)}^{2}}=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x-1=0 \\\\ & {{\\left( x-2 \\right)}^{2}}=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & x=1 \\\\ & x=2 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 \u0111\u00e1p \u00e1n ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1$ v\u00e0 $2$. <\/span><\/span> "}]}],"id_ques":576},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["0"],["1"],["3"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai6/lv3/img\/9.jpg' \/><\/center> Bi\u1ebft $(x+1)(6x^2+2x)$$+(x-1)(6x^2+2x)=0$, gi\u00e1 tr\u1ecb $\\left[ \\begin{align} & x= \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\\\ & x=-\\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}} \\\\ \\end{align} \\right.$ ","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 v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed: \u0111\u1eb7t $(6x^2+2x)$ l\u00e0m nh\u00e2n t\u1eed chung.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Ti\u1ebfp t\u1ee5c ph\u00e2n t\u00edch $6x^2+2x$ b\u1eb1ng c\u00e1ch \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/> <b> B\u01b0\u1edbc 3: <\/b>T\u00ecm $x$. N\u1ebfu $a.b = 0$ th\u00ec$ a = 0$ ho\u1eb7c $b = 0$. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3:<br\/> $ (x+1)(6{{x}^{2}}+2x)+(x-1)(6{{x}^{2}}+2x)=0 \\\\ \\Leftrightarrow (6{{x}^{2}}+2x)\\left( x+1+x-1 \\right)=0 \\\\ \\Leftrightarrow (6{{x}^{2}}+2x).2x=0 \\\\ \\Leftrightarrow 2x\\left( 3x+1 \\right)2x=0 \\\\ \\Leftrightarrow 4{{x}^{2}}\\left( 3x+1 \\right)=0 \\Leftrightarrow \\left[ \\begin{aligned} & 4{{x}^{2}}=0 \\\\ & 3x+1=0 \\\\ \\end{aligned} \\right. \\Leftrightarrow \\left[ \\begin{aligned} & x=0 \\\\ & x=\\dfrac{-1}{3} \\\\ \\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 $0$ v\u00e0 $\\dfrac{1}{3}$. <\/span><\/span> "}]}],"id_ques":577},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai6/lv3/img\/8.jpg' \/><\/center> Ph\u00e2n t\u00edch \u0111a th\u1ee9c $a-b+(b-a)^2$ th\u00e0nh nh\u00e2n t\u1eed, ta \u0111\u01b0\u1ee3c:","select":["A. $(a-b)(1-a+b)$ ","B. $(a-b)(1-a-b)$ ","C. $(a-b)(1+a+b)$","D. $(a-b)(1+a-b)$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> \u0110\u1eb7t $a-b$ ho\u1eb7c $b-a$ l\u00e0 nh\u00e2n t\u1eed chung \u0111\u1ec3 ph\u00e2n t\u00edch .<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'>Ta c\u00f3:<br\/> $\\begin{align} & a-b+{{(b-a)}^{2}} \\\\ & =\\left( a-b \\right)+{{\\left( a-b \\right)}^{2}} \\\\ & =\\left( a-b \\right)\\left( 1+a-b \\right) \\\\ \\end{align}$<span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span>","column":2}]}],"id_ques":578},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai6/lv3/img\/4.jpg' \/><\/center> Ph\u00e2n t\u00edch \u0111a th\u1ee9c $(64a^3+125b^3)+5b(16a^2-25b^2)$ th\u00e0nh nh\u00e2n t\u1eed, ta \u0111\u01b0\u1ee3c:","select":["A. $16a^2(4a-5b)$ ","B. $a^2(4a+5b)$ ","C. $16a^2(4a+5b)$","D. $a^2(4a-5b)$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>Ta c\u00f3 $64a^3+125b^3=(4a)^3+(5b)^3$, $16a^2-25b^2=(4a)^2-(5b)^2$.<br\/> Ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed theo h\u1eb1ng \u0111\u1eb3ng th\u1ee9c v\u00e0 \u0111\u1eb7t nh\u00e2n t\u1eed chung. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'>Ta c\u00f3:<br\/> $\\begin{align} & (64{{a}^{3}}+125{{b}^{3}})+5b(16{{a}^{2}}-25{{b}^{2}}) \\\\ & =\\left[ {{\\left( 4a \\right)}^{3}}+{{\\left( 5b \\right)}^{3}} \\right]+5b\\left[ {{\\left( 4a \\right)}^{2}}-{{\\left( 5b \\right)}^{2}} \\right] \\\\ & =\\left( 4a+5b \\right)\\left( 16{{a}^{2}}-20ab+25{{b}^{2}} \\right)+5b\\left( 4a-5b \\right)\\left( 4a+5b \\right) \\\\ & =\\left( 4a+5b \\right)\\left[ \\left( 16{{a}^{2}}-20ab+25{{b}^{2}} \\right)+5b\\left( 4a-5b \\right) \\right] \\\\ & =\\left( 4a+5b \\right)\\left( 16{{a}^{2}}-20ab+25{{b}^{2}}+20ab-25{{b}^{2}} \\right) \\\\ & =\\left( 4a+5b \\right)16{{a}^{2}} \\\\ \\end{align}$<span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span>","column":2}]}],"id_ques":579},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai6/lv3/img\/2.jpg' \/><\/center> \u0110a th\u1ee9c $A=5x^n+10x^{n+2}$ (V\u1edbi $n > 0$ v\u00e0 $n \\in \\mathbb{N}$) c\u00f3 gi\u00e1 tr\u1ecb b\u1eb1ng $0$ khi:","select":["A. $x=\\pm \\dfrac{1}{2}$ ","B. $x=\\pm \\dfrac{1}{\\sqrt{2}}$ ","C. $x = 0; \\pm 2$","D. $x = 0$"],"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 theo ph\u01b0\u01a1ng ph\u00e1p \u0111\u1eb7t nh\u00e2n t\u1eed chung. <br\/> <b> B\u01b0\u1edbc 2:<\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 A = 0 <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ecm $x$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'>Ta c\u00f3 :<br\/> $\\begin{align} & A=5{{x}^{n}}+10{{x}^{n+2}} \\\\ & =5{{x}^{n}}\\left( 1+2{{x}^{2}} \\right) \\\\ & Do\\,\\,1+2{{x}^{2}}\\,\\,>\\,0 (\\forall x), \\Rightarrow {{x}^{n}}=0\\Leftrightarrow x=0 \\\\ \\end{align}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span>","column":2}]}],"id_ques":580}],"lesson":{"save":0,"level":3}}