đang tải bài tập bài
{"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}}

Điểm của bạn.

Câu hỏi này theo dạng chọn đáp án đúng, sau khi đọc xong câu hỏi, bạn bấm vào một trong số các đáp án mà chương trình đưa ra bên dưới, sau đó bấm vào nút gửi để kiểm tra đáp án và sẵn sàng chuyển sang câu hỏi kế tiếp

Trả lời đúng trong khoảng thời gian quy định bạn sẽ được + số điểm như sau:
Trong khoảng 1 phút đầu tiên + 5 điểm
Trong khoảng 1 phút -> 2 phút + 4 điểm
Trong khoảng 2 phút -> 3 phút + 3 điểm
Trong khoảng 3 phút -> 4 phút + 2 điểm
Trong khoảng 4 phút -> 5 phút + 1 điểm

Quá 5 phút: không được cộng điểm

Tổng thời gian làm mỗi câu (không giới hạn)

Điểm của bạn.

Bấm vào đây nếu phát hiện có lỗi hoặc muốn gửi góp ý