{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai5/lv3/img\/5.jpg' \/><\/center>Cho $x+y=1$. Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $P=2(x^3+y^3)-3(x^2+y^2)$ l\u00e0 $- 1$. <b> \u0110\u00fang <\/b> hay <b> sai <\/b>?","select":["\u0110\u00fang","Sai"],"hint":" Khai tri\u1ec3n bi\u1ec3u th\u1ee9c \u0111\u00e3 cho theo h\u1eb1ng \u0111\u1eb3ng th\u1ee9c \u0111\u00e3 h\u1ecdc, \u0111\u01b0a v\u1ec1 k\u1ebft qu\u1ea3 ch\u1ec9 ch\u1ee9a $x + y$.<br\/> Thay $x+y=1$ v\u00e0o bi\u1ec3u th\u1ee9c \u0111\u1ec3 t\u00ednh. ","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$ P=2({{x}^{3}}+{{y}^{3}})-3({{x}^{2}}+{{y}^{2}}) $<br\/>$ =2\\left( x+y \\right)\\left( {{x}^{2}}-xy+{{y}^{2}} \\right)$$-3{{x}^{2}}-3{{y}^{2}} $<br\/>$ =2.1.\\left( {{x}^{2}}-xy+{{y}^{2}} \\right)$$-3{{x}^{2}}-3{{y}^{2}}$(V\u00ec $x+y=1$)<br\/>$ =2{{x}^{2}}-2xy+2{{y}^{2}}$$-3{{x}^{2}}-3{{y}^{2}} $<br\/>$ =-{{x}^{2}}-2xy-{{y}^{2}} $<br\/>$ =-{{\\left( x+y \\right)}^{2}} $ <br\/>V\u1edbi $x + y = 1$ th\u00ec $ P = -(x+y)^2=-1^2=-1$<br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0: \u0110\u00fang.<\/span>","column":2}]}],"id_ques":521},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["-18"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai5/lv3/img\/3.jpg' \/><\/center> H\u1ec7 s\u1ed1 c\u1ee7a $x^2$ trong \u0111a th\u1ee9c $A=(x-3)^3-(x+3)^3$ l\u00e0 _input_ ","hint":"R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $A$, t\u1eeb \u0111\u00f3 t\u00ecm h\u1ec7 s\u1ed1 \u0111i v\u1edbi $x^2$. ","explain":"<span class='basic_left'>Ta c\u00f3:<br\/> $ A={{(x-3)}^{3}}-{{(x+3)}^{3}} $<br\/>$ ={{x}^{3}}-9{{x}^{2}}+27x-27$$-\\left( {{x}^{3}}+9{{x}^{2}}+27x+27 \\right) $<br\/>$ ={{x}^{3}}-9{{x}^{2}}+27x-27-{{x}^{3}}$$-9{{x}^{2}}-27x-27 $<br\/>$ =\\left( {{x}^{3}}-{{x}^{3}} \\right)+\\left( -9{{x}^{2}}-9{{x}^{2}} \\right)$$+\\left( 27x-27x \\right)+\\left( -27-27 \\right) $<br\/>$ =-18{{x}^{2}}-54 $.<br\/> H\u1ec7 s\u1ed1 \u0111i v\u1edbi $x^2$ l\u00e0 $- 18$. <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-18$. <\/span><\/span> "}]}],"id_ques":522},{"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/bai5/lv3/img\/1.png' \/><\/center> Cho $x+y=1$ v\u00e0 $xy=-1$. Gi\u00e1 tr\u1ecb c\u1ee7a $x^3+y^3$ 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> Khai tri\u1ec3n bi\u1ec3u th\u1ee9c \u0111\u00e3 cho theo h\u1eb1ng \u0111\u1eb3ng th\u1ee9c t\u1ed5ng hai l\u1eadp ph\u01b0\u01a1ng, \u0111\u01b0a v\u1ec1 k\u1ebft qu\u1ea3 ch\u1ec9 ch\u1ee9a $x + y$ v\u00e0 $xy$ <br\/> <b> B\u01b0\u1edbc 2:<\/b> Thay $x+y=1$ v\u00e0 $xy=-1$ v\u00e0o bi\u1ec3u th\u1ee9c \u0111\u1ec3 t\u00ednh.<br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'>Ta c\u00f3 :<br\/> $x^3+y^3=(x+y)(x^2-xy+y^2)$$=(x+y)[(x+y)^2-3xy]$.<br\/> Thay $x+y=1$ v\u00e0 $xy=-1$ v\u00e0o bi\u1ec3u th\u1ee9c \u0111\u00e3 r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c: <br\/> $(x+y)[(x+y)^2-3xy]$$=1.[1^2-3.(-1)]=1.4=4 $.<span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span>","column":2}]}],"id_ques":523},{"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/bai5/lv3/img\/2.jpg' \/><\/center> Bi\u1ec3u th\u1ee9c $B=(2x-1)(4x^2+2x+1)$$-4x(2x^2-3)$ c\u00f3 gi\u00e1 tr\u1ecb b\u1eb1ng $5$ khi:","select":["A. $x=\\dfrac{1}{4}$ ","B. $x=-\\dfrac{1}{2}$ ","C. $x=-\\dfrac{1}{4}$","D. $x=\\dfrac{1}{2}$"],"hint":"R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $B$.<br\/>Cho $B = 5$ v\u00e0 t\u00ecm $x$. ","explain":"<span class='basic_left'>Ta c\u00f3:<br\/> $ B=(2x-1)(4{{x}^{2}}+2x+1)-4x(2{{x}^{2}}-3) $<br\/>$ ={{\\left( 2x \\right)}^{3}}-{{1}^{3}}-8{{x}^{3}}+12x $<br\/>$ =8{{x}^{3}}-1-8{{x}^{3}}+12x $<br\/>$ =-1+12x $ <br\/>$B = 5$<br\/>$\\begin{align} &\\Rightarrow -1+12x=5 \\\\ &\\Leftrightarrow 12x=5+1 \\\\ &\\Leftrightarrow 12x=6 \\\\ & \\Leftrightarrow x=\\dfrac{1}{2} \\\\ \\end{align}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span>","column":2}]}],"id_ques":524},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["3367"],["1000"]]],"list":[{"point":10,"width":60,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai5/lv3/img\/4.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $\\left( \\dfrac{1}{4}xy-\\dfrac{1}{5} \\right)$$\\left( \\dfrac{1}{16}{{x}^{2}}{{y}^{2}}+\\dfrac{1}{20}xy+\\dfrac{1}{25} \\right)$ v\u1edbi $x = 2; y = 3$ l\u00e0 <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c.<br\/> <b>B\u01b0\u1edbc 2:<\/b> Thay $x = 2; y = 3$ 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\/> $\\left( \\dfrac{1}{4}xy-\\dfrac{1}{5} \\right)$$\\left( \\dfrac{1}{16}{{x}^{2}}{{y}^{2}}+\\dfrac{1}{20}xy+\\dfrac{1}{25} \\right)$$={{\\left( \\dfrac{1}{4}xy \\right)}^{3}}-{{\\left( \\dfrac{1}{5} \\right)}^{3}}$ <br\/> Thay $x = 2; y = 3$ v\u00e0o bi\u1ec3u th\u1ee9c \u0111\u00e3 r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c: <br\/> ${{\\left( \\dfrac{1}{4}xy \\right)}^{3}}-{{\\left( \\dfrac{1}{5} \\right)}^{3}}$$={{\\left( \\dfrac{1}{4}.2.3 \\right)}^{3}}-\\dfrac{1}{125}$$={{\\left( \\dfrac{3}{2} \\right)}^{3}}-\\dfrac{1}{125}$$=\\dfrac{27}{8}-\\dfrac{1}{125}=\\dfrac{3367}{1000} $.<\/span> "}]}],"id_ques":525},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["-1"],["2"]]],"list":[{"point":10,"width":40,"type_input":"","input_hint":["frac","sqr"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai5/lv3/img\/3.jpg' \/><\/center>Bi\u1ebft $(x+1)^3-(x-1)^3-6(x-1)^2$$=-10$, gi\u00e1 tr\u1ecb $x$ l\u00e0 <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","hint":"S\u1eed d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c \u0111\u1ec3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c v\u1ebf tr\u00e1i v\u00e0 t\u00ecm $x$.","explain":"<span class='basic_left'>Ta c\u00f3:<br\/> ${{(x+1)}^{3}}-{{(x-1)}^{3}}$$-6{{(x-1)}^{2}}=-10 $<br\/>$\\Leftrightarrow {{x}^{3}}+3{{x}^{2}}+3x+1-\\left( {{x}^{3}}-3{{x}^{2}}+3x-1 \\right)$$-6\\left( {{x}^{2}}-2x+1 \\right)=-10 $<br\/>$ \\Leftrightarrow{{x}^{3}}+3{{x}^{2}}+3x+1-{{x}^{3}}+3{{x}^{2}}-3x+1$$-6{{x}^{2}}+12x-6=-10 $<br\/>$ \\Leftrightarrow\\left( {{x}^{3}}-{{x}^{3}} \\right)+\\left( 3{{x}^{2}}+3{{x}^{2}}-6{{x}^{2}} \\right)$$+\\left( 3x-3x+12x \\right)+1+1-6=-10 $<br\/>$ \\Leftrightarrow12x-4=-10 $<br\/>$\\Leftrightarrow 12x=-10+4 $<br\/>$\\Leftrightarrow12x=-6 $<br\/>$\\Leftrightarrow x=\\dfrac{-1}{2} $<\/span> "}]}],"id_ques":526},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u + ho\u1eb7c - v\u00e0o \u00f4 tr\u1ed1ng \u0111\u1ec3 \u0111\u01b0\u1ee3c m\u1ed9t khai tri\u1ec3n \u0111\u00fang ","title_trans":"","temp":"fill_the_blank","correct":[[["-"],["+"],["+"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai5/lv3/img\/5.jpg' \/><\/center>$ 0,001{{x}^{3}}+\\dfrac{1}{64} $$=\\left( 0,01{{x}^{2}}\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}0,025x\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\dfrac{1}{16} \\right)\\left( 0,1x\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\dfrac{1}{4} \\right)$ ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/> <b>B\u01b0\u1edbc 1:<\/b> Ta c\u00f3 $ 0,001{{x}^{3}}+\\dfrac{1}{64} $$={{\\left( 0,1x \\right)}^{3}}+{{\\left( \\dfrac{1}{4} \\right)}^{3}}$<br\/> <b> B\u01b0\u1edbc 2: <\/b>Ta khai tri\u1ec3n theo h\u1eb1ng \u0111\u1eb3ng th\u1ee9c: $a^3+b^3$, t\u1eeb \u0111\u00f3 t\u00ecm ra c\u00e1c d\u1ea5u c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Ta c\u00f3:<br\/>$\\left( 0,001{{x}^{3}}+\\dfrac{1}{64} \\right)$$={{\\left( 0,1x \\right)}^{3}}+{{\\left( \\dfrac{1}{4} \\right)}^{3}}$$=\\left( 0,1x+\\dfrac{1}{4} \\right)\\left( 0,01{{x}^{2}}-0,025x+\\dfrac{1}{16} \\right)$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t c\u00e1c d\u1ea5u $-, +$ v\u00e0 $+$. <\/span><\/span> "}]}],"id_ques":527},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $a = \\frac{3}{2}; b = 4; c = 6$","B. $a = \\frac{3}{2}; b = 2; c = 4$","C. $a = \\frac{3}{2}; b = 6; c = 6$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai5/lv3/img\/2.jpg' \/><\/center> $ \\dfrac{27}{8}{{z}^{3}}-64{{x}^{3}} =(az- bx ) ( \\dfrac{9}{4}{{z}^{2}}+cxz +16{{x}^{2}} )$. $a,b,c =?$","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/>V\u1ebf tr\u00e1i l\u00e0 h\u1eb1ng \u0111\u1eb3ng th\u1ee9c:$a^3-b^3$.<br\/> Ta khai tri\u1ec3n \u0111\u1ec3 t\u00ecm c\u00e1c y\u1ebfu t\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Ta c\u00f3: <br\/> $\\dfrac{27}{8}{{z}^{3}}-64{{x}^{3}}$$={{\\left( \\dfrac{3}{2}z \\right)}^{3}}-{{\\left( 4x \\right)}^{3}}$$=\\left( \\dfrac{3}{2}z-4x \\right)\\left( \\dfrac{9}{4}{{z}^{2}}+6xz+16{{x}^{2}} \\right)$<\/span> "}]}],"id_ques":528},{"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/bai5/lv3/img\/4.jpg' \/><\/center> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $A=x\\left( x+2 \\right)\\left( x-2 \\right)$$-\\left( x-3 \\right)\\left( {{x}^{2}}+3x+9 \\right)$ ta \u0111\u01b0\u1ee3c:","select":["A. $x^2-4$ ","B. $x^3-12$ ","C. $-4x+27$","D. $x^3-2$"],"hint":"R\u00fat g\u1ecdn theo h\u1eb1ng \u0111\u1eb3ng th\u1ee9c hi\u1ec7u hai b\u00ecnh ph\u01b0\u01a1ng v\u00e0 hi\u1ec7u hai l\u1eadp ph\u01b0\u01a1ng.","explain":"<span class='basic_left'>Ta c\u00f3:<br\/> $ A=x\\left( x+2 \\right)\\left( x-2 \\right)$$-\\left( x-3 \\right)\\left( {{x}^{2}}+3x+9 \\right) $<br\/>$ =x\\left( {{x}^{2}}-4 \\right)-\\left( {{x}^{3}}-{{3}^{3}} \\right) $<br\/>$ ={{x}^{3}}-4x-{{x}^{3}}+27 $<br\/>$ =-4x+27 $<span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span>","column":2}]}],"id_ques":529},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai5/lv3/img\/3.jpg' \/><\/center>Khai tri\u1ec3n $\\dfrac{125}{64}{{a}^{6}}-{{b}^{6}}$ theo h\u1eb1ng \u0111\u1eb3ng th\u1ee9c ta \u0111\u01b0\u1ee3c:","select":["A. $\\left( \\dfrac{5}{4}{{a}^{3}}-{{b}^{3}} \\right)\\left( \\dfrac{25}{16}{{a}^{6}}+\\dfrac{5}{4}{{a}^{3}}{{b}^{3}}+{{b}^{6}} \\right)$ ","B. $\\left( \\dfrac{5}{4}{{a}^{2}}-{{b}^{2}} \\right)\\left( \\dfrac{25}{16}{{a}^{4}}+\\dfrac{5}{4}{{a}^{2}}{{b}^{2}}+{{b}^{4}} \\right)$ ","C. $\\left( \\dfrac{5}{4}{{a}^{2}}-{{b}^{2}} \\right)\\left( \\dfrac{25}{16}{{a}^{4}}-\\dfrac{5}{4}{{a}^{2}}{{b}^{2}}+{{b}^{4}} \\right)$","D. $\\left( \\dfrac{5}{4}{{a}}-{{b}} \\right)\\left( \\dfrac{25}{16}{{a}^{2}}+\\dfrac{5}{4}ab+{{b}^{2}} \\right)$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/> Khai tri\u1ec3n theo h\u1eb1ng \u0111\u1eb3ng th\u1ee9c:$a^3-b^3$.<br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'>Ta c\u00f3:<br\/> $\\dfrac{125}{64}{{a}^{6}}-{{b}^{6}}$$={{\\left( \\dfrac{5}{4}{{a}^{2}} \\right)}^{3}}-{{\\left( {{b}^{2}} \\right)}^{3}}$$=\\left( \\dfrac{5}{4}{{a}^{2}}-{{b}^{2}} \\right)$$\\left[ {{\\left( \\dfrac{5}{4}{{a}^{2}} \\right)}^{2}}+\\dfrac{5}{4}{{a}^{2}}{{b}^{2}}+{{\\left( {{b}^{2}} \\right)}^{2}} \\right]$$=\\left( \\dfrac{5}{4}{{a}^{2}}-{{b}^{2}} \\right)$$\\left( \\dfrac{25}{16}{{a}^{4}}+\\dfrac{5}{4}{{a}^{2}}{{b}^{2}}+{{b}^{4}} \\right)$<span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span>","column":2}]}],"id_ques":530}],"lesson":{"save":0,"level":3}}