{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Hai h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t hai \u1ea9n c\u00f9ng v\u00f4 nghi\u1ec7m th\u00ec t\u01b0\u01a1ng \u0111\u01b0\u01a1ng v\u1edbi nhau. Kh\u1eb3ng \u0111\u1ecbnh tr\u00ean \u0111\u00fang hay sai? ","select":["A. \u0110\u00fang","B. Sai"],"explain":" <span class='basic_left'> Hai h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t hai \u1ea9n c\u00f9ng v\u00f4 nghi\u1ec7m th\u00ec c\u00f3 c\u00f9ng t\u1eadp nghi\u1ec7m l\u00e0 $S=\\varnothing $ <br\/> Hai h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 c\u00f9ng t\u1eadp nghi\u1ec7m th\u00ec t\u01b0\u01a1ng \u0111\u01b0\u01a1ng v\u1edbi nhau <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A <\/span>","column":2}]}],"id_ques":241},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Cho ph\u01b0\u01a1ng tr\u00ecnh $2x-3y=1.$ T\u00ecm $b$ \u0111\u1ec3 c\u1eb7p s\u1ed1 $(10;b^2)$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh.","select":["A. $b=\\pm\\sqrt{\\dfrac{19}{3}}$ ","B. $b=\\pm\\sqrt{\\dfrac{21}{3}}$","C. $b=\\pm\\dfrac{361}{9}$","D. $b=\\pm\\dfrac{441}{9}$"],"explain":" <span class='basic_left'> Thay $x=10;y=b^2$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh $2x-3y=1$ ta \u0111\u01b0\u1ee3c: <br\/> $2.10-3.b^2=1$ <br\/> $\\Leftrightarrow 20-3b^2=1$ $\\Leftrightarrow b^2=\\dfrac{19}{3}$ $\\Leftrightarrow b=\\pm\\sqrt{\\dfrac{19}{3}}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A <\/span>","column":4}]}],"id_ques":242},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["8"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" Trong m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 $xOy,$ cho ba \u0111i\u1ec3m $A(1;2), B(2;5), C(3;k).$ T\u00ecm $k$ \u0111\u1ec3 ba \u0111i\u1ec3m $A, B, C$ th\u1eb3ng h\u00e0ng? <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> $k=$_input_ ","hint":"Ba \u0111i\u1ec3m th\u1eb3ng h\u00e0ng n\u1ebfu ch\u00fang c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng hay ch\u00fang c\u00f9ng l\u00e0 nghi\u1ec7m c\u1ee7a m\u1ed9t ph\u01b0\u01a1ng tr\u00ecnh","explain":" <span class='basic_left'> G\u1ecdi $(d): y=ax+b$ l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua ba \u0111i\u1ec3m $A,B,C,$ <br\/> Do $A, B, C$ thu\u1ed9c $(d)$ n\u00ean t\u1ecda \u0111\u1ed9 c\u1ee7a ch\u00fang th\u1ecfa m\u00e3n ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1eddng th\u1eb3ng, t\u1ee9c l\u00e0: $\\left\\{ \\begin{align}& a+b=2\\, (2)\\\\ & 2a+b=5\\, (2)\\\\ & 3a+b=k\\, (3)\\\\ \\end{align} \\right.$ <br\/> (1) $\\Leftrightarrow 2a+2b=4 \\Leftrightarrow 2a + b + b = 4$ (*) <br\/> Thay (2) v\u00e0o (*) ta \u0111\u01b0\u1ee3c: <br\/> $ 5+b=4 \\Leftrightarrow b = -1$ <br\/> Thay $b=-1$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh (1) ta \u0111\u01b0\u1ee3c: <br\/> $a - 1 = 2 \\Leftrightarrow a= 3$ <br\/> Thay $a = 3; b=-1$ v\u00e0o (3) ta \u0111\u01b0\u1ee3c: <br\/> $3.(-3) - 1 = k \\Leftrightarrow k = 8$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $8$ <\/span>"}]}],"id_ques":243},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"Cho h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{align}& a_1x+b_1y=c_1 \\\\ & a_2x+b_2y=c_2\\\\ \\end{align} \\right.$(*) <br\/> V\u1edbi $a_1; b_1 \\ne 0.$ H\u1ec7 v\u00f4 nghi\u1ec7m khi:","select":["A. $\\dfrac{{{a}_{1}}}{{{a}_{2}}}\\ne \\dfrac{{{b}_{1}}}{{{b}_{2}}}$ ","B. $\\dfrac{{{a}_{1}}}{{{a}_{2}}}=\\dfrac{{{b}_{1}}}{{{b}_{2}}}\\ne \\dfrac{{{c}_{1}}}{{{c}_{2}}}$","C. $\\dfrac{{{a}_{1}}}{{{a}_{2}}}=\\dfrac{{{b}_{1}}}{{{b}_{2}}}=\\dfrac{{{c}_{1}}}{{{c}_{2}}}$","D. $\\dfrac{{{b}_{1}}}{{{b}_{2}}}=\\dfrac{{{c}_{1}}}{{{c}_{2}}}$"],"explain":" <span class='basic_left'> H\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh (*) v\u00f4 nghi\u1ec7m n\u1ebfu hai \u0111\u01b0\u1eddng th\u1eb3ng $a_1x+b_1y=c_1$ v\u00e0 $a_2x+b_2y=c_2$ song song <br\/> $\\Leftrightarrow \\dfrac{{{a}_{1}}}{{{a}_{2}}}=\\dfrac{{{b}_{1}}}{{{b}_{2}}}\\ne \\dfrac{{{c}_{1}}}{{{c}_{2}}}$<br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B <\/span>","column":2}]}],"id_ques":244},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["2"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" Cho ba \u0111\u01b0\u1eddng th\u1eb3ng: $ (d_1): x+y-4=0 \\, (d_2): x=3\\, (d_3): \\dfrac{x}{6}+\\dfrac{y}{k}=1.$ N\u1ebfu ba \u0111\u01b0\u1eddng th\u1eb3ng \u0111\u1ed3ng quy th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $k$ l\u00e0_input_ ","hint":"Ba \u0111\u01b0\u1eddng th\u1eb3ng \u0111\u1ed3ng quy n\u1ebfu ch\u00fang c\u00f9ng \u0111i qua m\u1ed9t \u0111i\u1ec3m ","explain":" <span class='basic_left'> X\u00e9t h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{\\begin{align}& x+y-4=0 (1) \\\\ & x=3 \\end{align} \\right.$ <br\/> Thay $x= 3$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh (1) ta \u0111\u01b0\u1ee3c: <br\/> $3 + y - 4 = 0 \\Leftrightarrow y - 1 = 0 \\Leftrightarrow y = 1$ <br\/> Suy ra $(3;1)$ l\u00e0 nghi\u1ec7m c\u1ee7a h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh <br\/> T\u1ee9c l\u00e0 hai \u0111\u01b0\u1eddng th\u1eb3ng $(d_1)$ v\u00e0 $(d_2)$ c\u00f9ng \u0111i qua $(3;1)$ <br\/> \u0110\u1ec3 ba \u0111\u01b0\u1eddng th\u1eb3ng \u0111\u1ed3ng quy th\u00ec $(d_3)$ \u0111i qua $(3;1)$ <br\/> $\\Leftrightarrow \\dfrac{3}{6}+\\dfrac{1}{k}=1$ $\\Leftrightarrow \\dfrac{1}{k} = \\dfrac{1}{2}$ $\\Leftrightarrow k=2$<br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $2$ <\/span>"}]}],"id_ques":245},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"X\u00e1c \u0111\u1ecbnh $a,b$ \u0111\u1ec3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $ \\left\\{ \\begin{align} & ax + 8\\sqrt{3}y=20 \\\\ & 3\\sqrt{2}x+by=\\dfrac{9}{2} \\\\\\end{align} \\right.$ (*) c\u00f3 nghi\u1ec7m $\\left(\\sqrt{2};\\dfrac{\\sqrt{3}}{2}\\right)$","select":["A. $a=8\\sqrt{2};$ $b=\\sqrt{3}$ ","B. $a=8\\sqrt{2};$ $b=-\\sqrt{3}$","C. $a=4\\sqrt{2};$ $b=-\\sqrt{3}$","D. $a=4\\sqrt{2};$ $b=\\sqrt{3}$"],"explain":" <span class='basic_left'> Thay $x=\\sqrt{2};y=\\dfrac{\\sqrt{3}}{2}$ v\u00e0o h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh (*), ta \u0111\u01b0\u1ee3c: <br\/> $ \\left\\{ \\begin{align} & a\\sqrt{2} + 8\\sqrt{3}.\\dfrac{\\sqrt{3}}{2}=20 \\\\ & 3\\sqrt{2}.\\sqrt{2}+b\\dfrac{\\sqrt{3}}{2}=\\dfrac{9}{2} \\\\\\end{align} \\right.$ <br\/> $\\Leftrightarrow \\left\\{ \\begin{align} & a\\sqrt{2} + 12=20 \\\\ & 12+b\\sqrt{3}=9 \\\\\\end{align} \\right.$ <br\/> $ \\Leftrightarrow\\left\\{ \\begin{align} & a\\sqrt{2} =8 \\\\ & b\\sqrt{3}=-3 \\\\\\end{align} \\right.$ <br\/> $ \\left\\{ \\begin{align} & a=4\\sqrt{2} \\\\ & b=-\\sqrt{3} \\\\\\end{align} \\right.$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C <\/span>","column":2}]}],"id_ques":246},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["2"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" X\u00e1c \u0111\u1ecbnh $a$ \u0111\u1ec3 hai h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh sau t\u01b0\u01a1ng \u0111\u01b0\u01a1ng? <br\/> (I) $\\left\\{ \\begin{align}& x-y=1 \\\\ & 2x +y=2\\\\ \\end{align} \\right.;$ (II) $\\left\\{ \\begin{align}& ax-2y=2 \\\\ & x+ay=1 \\\\ \\end{align} \\right.$ <br\/><br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> $a = $ _input_ ","hint":"T\u00ecm nghi\u1ec7m c\u1ee7a h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh (I). Hai h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh t\u01b0\u01a1ng \u0111\u01b0\u01a1ng n\u1ebfu ch\u00fang c\u00f3 chung t\u1eadp nghi\u1ec7m","explain":" <span class='basic_left'>X\u00e9t h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh (I) $\\left\\{ \\begin{align}& x-y=1\\, (1) \\\\ & 2x +y=2\\, (2)\\\\ \\end{align} \\right.$ <br\/> Ph\u01b0\u01a1ng tr\u00ecnh (1) $\\Leftrightarrow 2x-2y=2 \\Leftrightarrow 2x+y-3y=2 $ (*) <br\/> Thay (2) v\u00e0o (*) ta \u0111\u01b0\u1ee3c: <br\/> $ 2-3y=2 \\Leftrightarrow y = 0$ <br\/> Thay $y = 0$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh (1) ta \u0111\u01b0\u1ee3c: <br\/> $2x = 2 \\Leftrightarrow x = 1$ <br\/> Suy ra $(1;0)$ l\u00e0 nghi\u1ec7m c\u1ee7a (I) <br\/> \u0110\u1ec3 hai h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh t\u01b0\u01a1ng \u0111\u01b0\u01a1ng th\u00ec $(1;0)$ c\u0169ng l\u00e0 nghi\u1ec7m c\u1ee7a (II) <br\/> Thay $x=1;y=0$ v\u00e0o (II) <br\/> $\\left\\{ \\begin{align}& a.1-2.0=2 (*) \\\\ & 1+a.0=1\\,\\text{(\u0111\u00fang)} \\\\ \\end{align} \\right.$ <br\/> $(*)\\Leftrightarrow a = 2$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $2$<\/span>"}]}],"id_ques":247},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"T\u00ecm c\u00f4ng th\u1ee9c nghi\u1ec7m nguy\u00ean t\u1ed5ng qu\u00e1t c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $5x+3y=2$ (1)","select":["A. $\\left\\{ \\begin{align}& x=3t+1 \\\\ & y=5t-1 \\\\ \\end{align} \\right.(t\\in \\mathbb{Z})$ ","B. $\\left\\{ \\begin{align}& x=3t-1 \\\\ & y=5t+1 \\\\ \\end{align} \\right.(t\\in \\mathbb{Z})$","C. $\\left\\{ \\begin{align}& x=3t-1 \\\\ & y=5t+5 \\\\ \\end{align} \\right.(t\\in \\mathbb{Z})$","D. $\\left\\{ \\begin{align}& x=3t+1 \\\\ & y=-5t-1 \\\\ \\end{align} \\right.(t\\in \\mathbb{Z})$"],"hint":"S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p t\u00ecm nghi\u1ec7m ri\u00eang (C\u00f3 th\u1ec3 em ch\u01b0a bi\u1ebft - trang 8 - sgk To\u00e1n 9)","explain":" <span class='basic_left'> Ta c\u00f3 $5.1 + 3.(-1) = 2$ <br\/> $\\Rightarrow (1; -1)$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $5x+3y=2$ <br\/> M\u00e0 $5 $ v\u00e0 $3$ l\u00e0 hai s\u1ed1 nguy\u00ean t\u1ed1 c\u00f9ng nhau <br\/> Suy ra c\u00f4ng th\u1ee9c nghi\u1ec7m nguy\u00ean c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh (1) l\u00e0: $\\left\\{ \\begin{align}& x=3t+1 \\\\ & y=-5t-1 \\\\ \\end{align} \\right.(t\\in \\mathbb{Z})$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D <\/span>","column":2}]}],"id_ques":248},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"T\u00ecm c\u00f4ng th\u1ee9c nghi\u1ec7m nguy\u00ean t\u1ed5ng qu\u00e1t c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $4x+11y=47$","select":["A. $\\left\\{ \\begin{align}& x=11t-9 \\\\ & y=4t-1 \\\\ \\end{align} \\right.(t\\in \\mathbb{Z})$ ","B. $\\left\\{ \\begin{align}& x=11t+9 \\\\ & y=-4t+1 \\\\ \\end{align} \\right.(t\\in \\mathbb{Z})$","C. $\\left\\{ \\begin{align}& x=-11t-9 \\\\ & y=-4t+1 \\\\ \\end{align} \\right.(t\\in \\mathbb{Z})$","D. $\\left\\{ \\begin{align}& x=11t+9 \\\\ & y=4t+1 \\\\ \\end{align} \\right.(t\\in \\mathbb{Z})$"],"hint":"S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p t\u00ecm nghi\u1ec7m ri\u00eang","explain":" <span class='basic_left'> Ta c\u00f3 $4.9+ 11.1 = 47$ <br\/> $\\Rightarrow (1; 1)$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $4x+11y=47$ <br\/> M\u00e0 $4 $ v\u00e0 $11$ l\u00e0 hai s\u1ed1 nguy\u00ean t\u1ed1 c\u00f9ng nhau <br\/> Suy ra c\u00f4ng th\u1ee9c nghi\u1ec7m nguy\u00ean c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh (1) l\u00e0: $\\left\\{ \\begin{align}& x=11t+9 \\\\ & y=-4t+1 \\\\ \\end{align} \\right.(t\\in \\mathbb{Z})$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B <\/span>","column":2}]}],"id_ques":249},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"M\u1ed9t x\u00ed nghi\u1ec7p c\u1ea7n chuy\u1ec3n 42 t\u1ea5n c\u00e1t t\u1edbi c\u00f4ng tr\u01b0\u1eddng x\u00e2y d\u1ef1ng b\u1eb1ng hai lo\u1ea1i xe 5 t\u1ea5n v\u00e0 9 t\u1ea5n. H\u1ecfi ph\u1ea3i huy \u0111\u1ed9ng bao nhi\u00eau xe m\u1ed7i lo\u1ea1i \u0111\u1ec3 l\u00e0m c\u00f4ng vi\u1ec7c n\u00e0y? (v\u1edbi \u0111i\u1ec1u ki\u1ec7n m\u1ed7i xe \u0111\u1ec1u ch\u1edf \u0111\u1ee7 tr\u1ecdng l\u01b0\u1ee3ng c\u1ee7a n\u00f3)","select":["A. $3$ v\u00e0 $3$ ","B. $12$ v\u00e0 $-2$","C. $-6$ v\u00e0 $8$","D. T\u1ea5t c\u1ea3 c\u00e1c \u0111\u00e1p \u00e1n tr\u00ean \u0111\u1ec1u \u0111\u00fang"],"explain":" <span class='basic_left'> G\u1ecdi s\u1ed1 xe tr\u1ecdng t\u1ea3i $5$ t\u1ea5n v\u00e0 $9$ t\u1ea5n c\u1ea7n d\u00f9ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $x$ v\u00e0 $y$ xe ($x,y\\in \\mathbb{N^*}$) <br\/> Do s\u1ed1 c\u00e1t c\u1ea7n v\u1eadn chuy\u1ec3n l\u00e0 $42$ t\u1ea5n n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh <br\/> $5x+9y=42$<br\/> Ta c\u00f3 $5.3 + 9.3 = 42$ <br\/> $\\Rightarrow (3; 3)$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $5x + 9y = 42$ <br\/> M\u00e0 $5$ v\u00e0 $9$ l\u00e0 hai s\u1ed1 nguy\u00ean t\u1ed1 c\u00f9ng nhau <br\/> Suy ra c\u00f4ng th\u1ee9c nghi\u1ec7m nguy\u00ean c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh (1) l\u00e0: $\\left\\{ \\begin{align}& x=9t+3 \\\\ & y=-5t+3 \\\\ \\end{align} \\right.(t\\in \\mathbb{Z})$<br\/> Do $x,y > 0$ n\u00ean <br\/> $\\left\\{ \\begin{align}& 9t + 3>0 \\\\ & -5t +3>0 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align}& t>\\dfrac{-1}{3} \\\\ & t<\\dfrac{3}{5} \\\\ \\end{align} \\right.$ <br\/> M\u00e0 $t\\in \\mathbb{Z}$$\\Rightarrow t=0$ <br\/> V\u1edbi $t = 0\\Rightarrow x=3; y=3$<br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A <\/span>","column":2}]}],"id_ques":250}],"lesson":{"save":0,"level":3}}