{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n c\u00e1c s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["10"],["0"],["-11"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"Gi\u1ea3i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{align} & x+4y=10 \\\\ & y-z=11 \\\\ & z+\\dfrac{x}{2}=-6 \\\\ \\end{align} \\right.$ <br\/><b> \u0110\u00e1p s\u1ed1:<\/b> H\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m duy nh\u1ea5t $(x;y;z)=$ (_input_;_input_;_input_)","hint":"Bi\u1ec3u di\u1ec5n $x$ v\u00e0 $y$ theo $z$ r\u1ed3i d\u00f9ng ph\u01b0\u01a1ng ph\u00e1p th\u1ebf \u0111\u1ec3 gi\u1ea3i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh","explain":"<span class='basic_left'>Ta x\u00e9t h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{align} & x+4y=10\\,\\,\\,\\,\\,\\left( 1 \\right) \\\\ & y-z=11\\,\\,\\,\\,\\,\\,\\,\\,\\left( 2 \\right) \\\\ & z+\\dfrac{x}{2}=-6\\,\\,\\,\\,\\,\\,\\left( 3 \\right) \\\\ \\end{align} \\right.$<br\/>T\u1eeb ph\u01b0\u01a1ng tr\u00ecnh (2), ta c\u00f3 $y=11+z$ (2\u2019)<br\/>T\u1eeb ph\u01b0\u01a1ng tr\u00ecnh (3), ta c\u00f3: $x=-12-2z$ (3\u2019)<br\/>Thay $y=11-z$ v\u00e0 $x=-12-2z$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh (1), ta c\u00f3: <br\/>$-12-2z+4\\left( 11+z \\right)=10\\Leftrightarrow 2z=-22\\Leftrightarrow z=-11$<br\/> Thay $z=-11$ v\u00e0o (2\u2019), ta c\u00f3 $y=11-11=0$<br\/>Thay $z=-11$ v\u00e0o (3\u2019), ta c\u00f3 $x=-12-2.(-11)=10$<br\/>V\u1eady h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 nghi\u1ec7m duy nh\u1ea5t l\u00e0 $(x;y;z)=(10;0;-11).$<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $10;0;-11$<br\/><\/span><\/span>"}]}],"id_ques":391},{"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":"<span class='basic_left'>Gi\u1ea3i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{align} & 2\\left( \\dfrac{1}{x}+\\dfrac{1}{2y} \\right)+3{{\\left( \\dfrac{1}{x}-\\dfrac{1}{2y} \\right)}^{2}}=9 \\\\ & \\dfrac{1}{x}+\\dfrac{1}{2y}-6{{\\left( \\dfrac{1}{x}-\\dfrac{1}{2y} \\right)}^{2}}=-3 \\\\ \\end{align} \\right.$<br\/> H\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m l\u00e0<\/span>","select":["A. $\\left( \\dfrac{1}{4};1 \\right)$ v\u00e0 $\\left( \\dfrac{1}{2};\\dfrac{1}{2} \\right)$","B. $\\left( 1;\\dfrac{1}{4} \\right)$ v\u00e0 $\\left( -\\dfrac{1}{2};-\\dfrac{1}{2} \\right)$","C. $\\left( 1;\\dfrac{1}{4} \\right)$ v\u00e0 $\\left( \\dfrac{1}{2};\\dfrac{1}{2} \\right)$","D. $\\left( 1;\\dfrac{1}{4} \\right)$ v\u00e0 $\\left( \\dfrac{3}{2};\\dfrac{1}{2} \\right)$ "],"hint":"D\u00f9ng ph\u01b0\u01a1ng ph\u00e1p \u0111\u1eb7t \u1ea9n ph\u1ee5 \u0111\u1ec3 gi\u1ea3i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $x,y \\ne 0$<br\/>\u0110\u1eb7t $u=\\dfrac{1}{x}+\\dfrac{1}{2y};v={{\\left( \\dfrac{1}{x}-\\dfrac{1}{2y} \\right)}^{2}}$ , ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{align} & 2u+3v=9 \\\\ & u-6v=-3 \\\\ \\end{align} \\right.$<br\/>$\\Leftrightarrow\\left\\{ \\begin{align} & 4u+6v=18 \\\\ & u-6v=-3 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & 5u=15 \\\\ & u-6v=-3 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & u=3 \\\\ & v=1 \\\\ \\end{align} \\right.$<br\/>Suy ra $\\left\\{ \\begin{align} & \\dfrac{1}{x}+\\dfrac{1}{2y}=3 \\\\ & {{\\left( \\dfrac{1}{x}-\\dfrac{1}{2y} \\right)}^{2}}=1 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & \\dfrac{1}{x}+\\dfrac{1}{2y}=3 \\\\ & \\left| \\dfrac{1}{x}-\\dfrac{1}{2y} \\right|=1 \\\\ \\end{align} \\right.$<br\/>$\\Leftrightarrow \\left\\{ \\begin{align} & \\dfrac{1}{x}=3-\\dfrac{1}{2y} \\\\ & \\left| \\dfrac{1}{x}-\\dfrac{1}{2y} \\right|=1 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & \\dfrac{1}{x}=3-\\dfrac{1}{2y} \\\\ & \\left| 3-\\dfrac{1}{2y}-\\dfrac{1}{2y} \\right|=1 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & \\dfrac{1}{x}=3-\\dfrac{1}{2y}\\,\\,\\,\\,\\left( 1 \\right) \\\\ & \\left| 3-\\dfrac{1}{y} \\right|=1\\,\\,\\,\\,\\,\\,\\,\\,\\left( 2 \\right) \\\\ \\end{align} \\right.$ <br\/>(2) $\\Leftrightarrow \\left[ \\begin{align} & 3-\\dfrac{1}{y}=1 \\\\ & 3-\\dfrac{1}{y}=-1 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left[ \\begin{align} & y=\\dfrac{1}{2} \\\\ & y=\\dfrac{1}{4} \\\\ \\end{align} \\right.$(th\u1ecfa m\u00e3n)<br\/>Thay $y=\\dfrac{1}{2}$ v\u00e0o (1), ta c\u00f3 $\\dfrac{1}{x}=2\\Leftrightarrow x=\\dfrac{1}{2}$ <br\/>Thay $y=\\dfrac{1}{4}$ v\u00e0o (1), ta c\u00f3 $\\dfrac{1}{x}=1\\Leftrightarrow x=1$ <br\/>V\u1eady h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 hai nghi\u1ec7m l\u00e0 $\\left( \\dfrac{1}{2};\\dfrac{1}{2} \\right)$ v\u00e0 $\\left( 1;\\dfrac{1}{4} \\right)$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":2}]}],"id_ques":392},{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n d\u1ea1ng $(x;y)$ \u00f4 tr\u1ed1ng","title_trans":"V\u00ed d\u1ee5: H\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m l\u00e0 $(2;4)$ th\u00ec ta \u0111i\u1ec1n l\u00e0 $2;4$ v\u00e0o \u00f4 tr\u1ed1ng","temp":"fill_the_blank_random","correct":[[["3;2"],["-5;2"]]],"list":[{"point":10,"width":100,"content":"","type_input":"","ques":"<span class='basic_left'>Gi\u1ea3i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{align} & \\left| x+1 \\right|+\\left| y-1 \\right|=5 \\\\ & \\left| x+1 \\right|=4y-4 \\\\ \\end{align} \\right.$<br\/><b> \u0110\u00e1p s\u1ed1:<\/b> H\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m hai nghi\u1ec7m l\u00e0 (_input_) v\u00e0 (_input_)","hint":"T\u1eeb ph\u01b0\u01a1ng tr\u00ecnh th\u1ee9 hai, ta x\u00e9t d\u1ea5u c\u1ee7a $y-1$ r\u1ed3i t\u1eeb \u0111\u00f3 gi\u1ea3i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh v\u1edbi hai \u1ea9n l\u00e0 $|x-1|$ v\u00e0 $y$","explain":"<span class='basic_left'>T\u1eeb ph\u01b0\u01a1ng tr\u00ecnh th\u1ee9 hai c\u1ee7a h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh, ta c\u00f3 $4y-4\\ge 0$ <br\/>Suy ra $y-1\\ge 0\\Rightarrow \\left| y-1 \\right|=y-1$ <br\/>Khi \u0111\u00f3 h\u1ec7 t\u01b0\u01a1ng \u0111\u01b0\u01a1ng v\u1edbi h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{align} & \\left| x+1 \\right|+y-1=5\\,\\, \\\\ & \\left| x+1 \\right|=4y-4\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & \\left| x+1 \\right|+y=6 \\\\ & \\left| x+1 \\right|-4y=-4 \\\\ \\end{align} \\right.$ <br\/>Tr\u1eeb t\u1eebng v\u1ebf ph\u01b0\u01a1ng tr\u00ecnh th\u1ee9 nh\u1ea5t cho t\u1eebng v\u1ebf ph\u01b0\u01a1ng tr\u00ecnh th\u1ee9 hai, ta c\u00f3<br\/>$\\left\\{ \\begin{align} & 5y=10 \\\\ & \\left| x+1 \\right|-4y=-4 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & y=2 \\\\ & \\left| x+1 \\right|=4 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{aligned} & y=2 \\\\ & \\left[ \\begin{aligned} & x+1=4 \\\\ & x+1=-4 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned} \\right.$ <br\/>$\\Leftrightarrow \\left\\{ \\begin{aligned} & y=2 \\\\ & \\left[ \\begin{aligned} & x=3 \\\\ & x=-5 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned} \\right.$ <br\/>V\u1eady h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 $\\left( 3;2 \\right)$ v\u00e0 $\\left( -5;2 \\right)$ <br\/><span class='basic_pink'>V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng \u0111\u1ec3 \u0111\u01b0\u1ee3c \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 $(3;2)$ v\u00e0 $(-5;2).$<br\/><\/span><\/span>"}]}],"id_ques":393},{"time":24,"part":[{"title":"\u0110i\u1ec1n c\u00e1c s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["1"],["-1"],["-3"],["-5"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'>Cho h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{align} & mx+2y=m+1 \\\\ & 2x+my=2m-1 \\\\ \\end{align} \\right.$ v\u1edbi $m$ l\u00e0 tham s\u1ed1<br\/>T\u00ecm c\u00e1c gi\u00e1 tr\u1ecb nguy\u00ean c\u1ee7a $m$ \u0111\u1ec3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m duy nh\u1ea5t l\u00e0 c\u00e1c s\u1ed1 nguy\u00ean <br\/><b> \u0110\u00e1p s\u1ed1:<\/b> $m \\in ${_input_;_input_;_input_;_input_}","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span> <br\/> B\u01b0\u1edbc 1: T\u00ecm $m$ \u0111\u1ec3 h\u1ec7 c\u00f3 nghi\u1ec7m duy nh\u1ea5t $(x;y).$<br\/>B\u01b0\u1edbc 2: Gi\u1ea3i h\u1ec7 t\u00ecm nghi\u1ec7m $(x;y)$ theo tham s\u1ed1 $m.$ Vi\u1ebft $x,$ $y$ d\u01b0\u1edbi d\u1ea1ng $a+\\dfrac{k}{f\\left( m \\right)}$ trong \u0111\u00f3 $a$ v\u00e0 $k$ l\u00e0 s\u1ed1 nguy\u00ean<br\/>B\u01b0\u1edbc 3: T\u00ecm $m$ \u0111\u1ec3 $x, y$ nguy\u00ean: $f(m)$ l\u00e0 \u01b0\u1edbc nguy\u00ean c\u1ee7a $k.$ So s\u00e1nh v\u1edbi \u0111i\u1ec1u ki\u1ec7n \u1edf b\u01b0\u1edbc 1 v\u00e0 k\u1ebft lu\u1eadn.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>T\u1eeb ph\u01b0\u01a1ng tr\u00ecnh th\u1ee9 nh\u1ea5t, bi\u1ec3u di\u1ec5n $y$ theo $x,$ ta \u0111\u01b0\u1ee3c h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh<br\/>$\\left\\{ \\begin{align} & y=\\dfrac{1}{2}\\left( -mx+m+1 \\right) \\\\ & 2x+my=2m-1 \\\\ \\end{align} \\right.$<br\/> $\\Leftrightarrow \\left\\{ \\begin{align} & y=\\dfrac{1}{2}\\left( -mx+m+1 \\right) \\\\ & 2x+\\dfrac{1}{2}m\\left( -mx+m+1 \\right)=2m-1 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & y=\\dfrac{1}{2}\\left( -mx+m+1 \\right) \\\\ & 4x-{{m}^{2}}x+{{m}^{2}}+m=4m-2 \\\\\\end{align} \\right.$<br\/> $\\Leftrightarrow \\left\\{ \\begin{align} & y=\\dfrac{1}{2}\\left( -mx+m+1 \\right) \\\\ & \\left( 4-{{m}^{2}} \\right)x+{{m}^{2}}-3m+2=0 \\\\ \\end{align} \\right.$ $\\Leftrightarrow \\left\\{ \\begin{align} & y=\\dfrac{1}{2}\\left( -mx+m+1 \\right) \\\\ & \\left( {{m}^{2}}-4 \\right)x=\\left( m-2 \\right)\\left( m-1 \\right) \\\\ \\end{align} \\right.$ (*)<br\/>H\u1ec7 \u0111\u00e3 cho c\u00f3 nghi\u1ec7m duy nh\u1ea5t $\\Leftrightarrow {{m}^{2}}-4\\ne 0\\Leftrightarrow m\\ne \\pm 2.$ <br\/>Khi \u0111\u00f3 $\\left( * \\right)\\Leftrightarrow \\left\\{ \\begin{aligned} & y=\\dfrac{1}{2}\\left( -mx+m+1 \\right) \\\\ & x=\\dfrac{m-1}{m+2} \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & y=\\dfrac{2m+1}{m+2} \\\\ & x=\\dfrac{m-1}{m+2} \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x=1-\\dfrac{3}{m+2} \\\\ & y=2-\\dfrac{3}{m+2} \\\\ \\end{aligned} \\right.$ <br\/>\u0110\u1ec3 $ x,$ $y$ nguy\u00ean th\u00ec $\\dfrac{3}{m+2}$ nguy\u00ean. Suy ra $ m+2$ l\u00e0 \u01b0\u1edbc nguy\u00ean c\u1ee7a $3$<br\/>Ta c\u00f3 b\u1ea3ng gi\u00e1 tr\u1ecb sau:<br\/><table> <tr> <td>$m+2$<\/td> <td>$1$<\/td> <td>$-1$ <\/td> <td>$3$<\/td><td>$-3$<\/td> <\/tr> <tr> <td>$m$<\/td> <td>$-1$<\/td> <td>$-3$<\/td> <td>$1$<\/td> <td>$-5$<\/td><\/tr> <\/table><br\/>C\u00e1c gi\u00e1 tr\u1ecb tr\u00ean c\u1ee7a $m$ \u0111\u1ec1u th\u1ecfa m\u00e3n. <br\/>V\u1eady v\u1edbi $m\\in \\text{ }\\!\\!\\{\\!\\!\\text{ }\\pm 1;-3;-5\\text{ }\\!\\!\\}\\!\\!\\text{ }$ th\u00ec h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m duy nh\u1ea5t $(x;y)$ th\u1ecfa m\u00e3n $x, y$ nguy\u00ean.<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1;-1;-3;-5$<\/span><\/span>"}]}],"id_ques":394},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["0"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"Cho h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{align} & x+my=m+1 \\\\ & mx+y=3m-1 \\\\ \\end{align} \\right.$ <br\/>T\u00ecm $m$ \u0111\u1ec3 h\u1ec7 c\u00f3 nghi\u1ec7m duy nh\u1ea5t $(x;y)$ sao cho t\u00edch $xy$ nh\u1ecf nh\u1ea5t. <br\/><b> \u0110\u00e1p s\u1ed1:<\/b> $m=$_input_","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span> <br\/> B\u01b0\u1edbc 1: T\u00ecm $m$ \u0111\u1ec3 h\u1ec7 c\u00f3 nghi\u1ec7m duy nh\u1ea5t $(x;y).$<br\/>B\u01b0\u1edbc 2: Gi\u1ea3i h\u1ec7 t\u00ecm nghi\u1ec7m $(x;y)$ theo tham s\u1ed1 $m.$ T\u00ednh t\u00edch $xy$<br\/>B\u01b0\u1edbc 3: Bi\u1ebfn \u0111\u1ed5i t\u00edch $xy$ v\u1ec1 d\u1ea1ng $A^2 +b$ r\u1ed3i \u0111\u00e1nh gi\u00e1 t\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t.<br\/>B\u01b0\u1edbc 4: T\u00ecm $m$ \u0111\u1ec3 d\u1ea5u ''$=$'' x\u1ea3y ra. So s\u00e1nh v\u1edbi \u0111i\u1ec1u ki\u1ec7n \u1edf b\u01b0\u1edbc 1 v\u00e0 k\u1ebft lu\u1eadn.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Bi\u1ec3u di\u1ec5n $x$ theo $y$ t\u1eeb ph\u01b0\u01a1ng tr\u00ecnh th\u1ee9 nh\u1ea5t c\u1ee7a h\u1ec7, ta c\u00f3:<br\/>$ \\left\\{ \\begin{align} & x=-my+m+1 \\\\ & m\\left( -my+m+1 \\right)+y=3m-1 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & x=-my+m+1 \\\\ & -{{m}^{2}}y+{{m}^{2}}+m+y=3m-1 \\\\ \\end{align} \\right.$<br\/>$\\Leftrightarrow \\left\\{ \\begin{align} & x=-my+m+1 \\\\ & \\left( {{m}^{2}}-1 \\right)y={{m}^{2}}-2m+1 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & x=-my+m+1 \\\\ & \\left( {{m}^{2}}-1 \\right)y={{\\left( m-1 \\right)}^{2}} \\\\ \\end{align} \\right.$ (*)<br\/>H\u1ec7 c\u00f3 nghi\u1ec7m duy nh\u1ea5t $\\Leftrightarrow {{m}^{2}}-1\\ne 0\\Leftrightarrow m\\ne \\pm 1$ <br\/>Khi \u0111\u00f3 $\\left( * \\right)\\Leftrightarrow \\left\\{ \\begin{align} & x=-my+m+1 \\\\ & y=\\dfrac{m-1}{m+1} \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & x=\\dfrac{3m+1}{m+1} \\\\ & y=\\dfrac{m-1}{m+1} \\\\ \\end{align} \\right.$ <br\/>$S=xy=\\dfrac{3m+1}{m+1}.\\dfrac{m-1}{m+1}=\\dfrac{\\left( 3m+1 \\right)\\left( m-1 \\right)}{{{\\left( m+1 \\right)}^{2}}}=\\dfrac{\\left[ 3\\left( m+1 \\right)-2 \\right].\\left[ \\left( m+1 \\right)-2 \\right]}{{{\\left( m+1 \\right)}^{2}}}$ <br\/>\u0110\u1eb7t $t=m+1$, ta c\u00f3 <br\/>$S=\\dfrac{\\left( 3t-2 \\right)\\left( t-2 \\right)}{{{t}^{2}}}=\\dfrac{3{{t}^{2}}-8t+4}{{{t}^{2}}}=3-\\dfrac{8}{t}+\\dfrac{4}{{{t}^{2}}}={{\\left( \\dfrac{2}{t}-2 \\right)}^{2}}-1$ <br\/>Ta c\u00f3 ${{\\left( \\dfrac{2}{t}-2 \\right)}^{2}}\\ge 0$ n\u00ean $S={{\\left( \\dfrac{2}{t}-2 \\right)}^{2}}-1\\ge -1$ <br\/>D\u1ea5u ''$=$'' x\u1ea3y ra khi $\\dfrac{2}{t}-2=0\\Leftrightarrow t=1\\Leftrightarrow m+1=1\\Leftrightarrow m=0$ (th\u1ecfa m\u00e3n) <br\/>V\u1eady $m=0$ th\u00ec h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m duy nh\u1ea5t $(x;y)$ th\u1ecfa m\u00e3n t\u00edch $xy$ nh\u1ecf nh\u1ea5t.<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0$<\/span><\/span>"}]}],"id_ques":395},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["3"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $\\sqrt{x+1}+\\sqrt[3]{x-2}=3$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m l\u00e0 $x=$_input_","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span> <br\/> B\u01b0\u1edbc 1: T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 ngh\u0129a<br\/>B\u01b0\u1edbc 2: \u0110\u1eb7t $u=\\sqrt{x+1};v=\\sqrt[3]{x-2}$ . L\u1eadp h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh v\u1edbi hai \u1ea9n $u$ v\u00e0 $v$<br\/>B\u01b0\u1edbc 3: Gi\u1ea3i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh, r\u1ed3i th\u1ebf tr\u1edf l\u1ea1i \u0111\u1ec3 t\u00ecm $x$ v\u00e0 $y$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x \\ge -1$<br\/>\u0110\u1eb7t $u=\\sqrt{x+1};v=\\sqrt[3]{x-2}$. Ta c\u00f3 $u+v=3$ (theo ph\u01b0\u01a1ng tr\u00ecnh ban \u0111\u1ea7u)<br\/>L\u1ea1i c\u00f3 ${{u}^{2}}-{{v}^{3}}=3$ <br\/>Ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh:$\\left\\{ \\begin{align} & u+v=3 \\\\ & {{u}^{2}}-{{v}^{3}}=3 \\\\ \\end{align} \\right.$<br\/> $\\Leftrightarrow \\left\\{ \\begin{align} & u=3-v \\\\ & {{\\left( 3-v \\right)}^{2}}-{{v}^{3}}=3 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & u=3-v \\\\ & 9-6v+{{v}^{2}}-{{v}^{3}}=3 \\\\ \\end{align} \\right.$ <br\/>$\\Leftrightarrow \\left\\{ \\begin{align} & u=3-v \\\\ & {{v}^{3}}-{{v}^{2}}+6v-6=0 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & u=3-v \\\\ & \\left( v-1 \\right)\\left( {{v}^{2}}+6 \\right)=0 \\\\ \\end{align} \\right.$<br\/>$\\Leftrightarrow \\left\\{ \\begin{align} & u=3-v \\\\ & v-1=0\\,\\,(\\text{v\u00ec}\\,{{v}^{2}}+6>0)\\\\ \\end{align} \\right.$ $\\Leftrightarrow \\left\\{ \\begin{align} & u=2 \\\\ & v=1 \\\\ \\end{align} \\right.$ <br\/>Suy ra $\\left\\{ \\begin{align} & \\sqrt{x+1}=2 \\\\ & \\sqrt[3]{x-2}=1 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & x+1=4 \\\\ & x-2=1 \\\\ \\end{align} \\right.$$\\Leftrightarrow x=3$ (th\u1ecfa m\u00e3n) <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m duy nh\u1ea5t $x=3$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $3$<\/span><\/span>"}]}],"id_ques":396},{"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":"Gi\u1ea3i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh$\\left\\{ \\begin{align} & {{x}^{2}}-{{y}^{2}}+2y=1 \\\\ & {{\\left( x+y \\right)}^{2}}-2x-2y=0 \\\\ \\end{align} \\right.$ <br\/>H\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m l\u00e0 ","select":["A.$\\left( \\dfrac{1}{2};-\\dfrac{1}{2} \\right)$ v\u00e0 $\\left( \\dfrac{1}{2};\\dfrac{3}{2} \\right)$","B. $\\left( -\\dfrac{1}{2};\\dfrac{1}{2} \\right)$ v\u00e0 $\\left( \\dfrac{3}{2};\\dfrac{1}{2} \\right)$","C. $\\left( -\\dfrac{1}{2};\\dfrac{1}{2} \\right)$ v\u00e0 $\\left( \\dfrac{1}{2};\\dfrac{3}{2} \\right)$","D. $\\left( -\\dfrac{1}{2};-\\dfrac{1}{2} \\right)$ v\u00e0 $\\left( \\dfrac{1}{2};\\dfrac{3}{2} \\right)$"],"hint":"\u0110\u01b0a ph\u01b0\u01a1ng tr\u00ecnh th\u1ee9 2 v\u1ec1 d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh t\u00edch","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\left\\{ \\begin{align} & {{x}^{2}}-{{y}^{2}}+2y=1\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 1 \\right) \\\\ & {{\\left( x+y \\right)}^{2}}-2x-2y=0\\,\\,\\,\\,\\,\\,\\,\\left( 2 \\right) \\\\ \\end{align} \\right.$<br\/>(2) $\\Leftrightarrow {{\\left( x+y \\right)}^{2}}-2\\left( x+y \\right)=0$<br\/> $\\begin{aligned} & \\Leftrightarrow \\left( x+y \\right)\\left( x+y-2 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x+y=0 \\\\ & x+y-2=0 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>+ V\u1edbi $x+y=0,$ suy ra $x=-y,$ thay v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh (1), ta c\u00f3<br\/>${{\\left( -y \\right)}^{2}}-{{y}^{2}}+2y=1\\Leftrightarrow 2y=1\\Leftrightarrow y=\\dfrac{1}{2}\\Rightarrow x=-\\dfrac{1}{2}$ <br\/>+ V\u1edbi $x+y-2=0$, suy ra $x=2-y.$ Thay v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh (1), ta c\u00f3:<br\/>${{\\left( 2-y \\right)}^{2}}-{{y}^{2}}+2y=1\\Leftrightarrow \\left( 4-4y+{{y}^{2}} \\right)-{{y}^{2}}+2y=1\\Leftrightarrow -2y=-3\\Leftrightarrow y=\\dfrac{3}{2}\\Rightarrow x=\\dfrac{1}{2}$<br\/>V\u1eady h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 hai nghi\u1ec7m l\u00e0 $\\left( -\\dfrac{1}{2};\\dfrac{1}{2} \\right)$ v\u00e0 $\\left( \\dfrac{1}{2};\\dfrac{3}{2} \\right)$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":2}]}],"id_ques":397},{"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":"<span class='basic_left'>Ba b\u1ea1n h\u1ecdc sinh v\u00e0o c\u1eeda h\u00e0ng mua \u0111\u1ed3 d\u00f9ng h\u1ecdc t\u1eadp. B\u1ea1n th\u1ee9 nh\u1ea5t mua $2$ c\u00e1i b\u00fat v\u00e0 $8$ quy\u1ec3n v\u1edf h\u1ebft $48000$ \u0111\u1ed3ng. B\u1ea1n th\u1ee9 hai mua $5$ quy\u1ec3n v\u1edf v\u00e0 $1$ quy\u1ec3n s\u1ed5 tay h\u1ebft $22000$ \u0111\u1ed3ng. B\u1ea1n th\u1ee9 ba mua $1$ c\u00e1i b\u00fat,$ 4$ quy\u1ec3n v\u1edf v\u00e0 $2$ quy\u1ec3n s\u1ed5 tay h\u1ebft $38 000$ \u0111\u1ed3ng. Gi\u00e1 ti\u1ec1n m\u1ed7i c\u00e1i b\u00fat, m\u1ed7i quy\u1ec3n v\u1edf v\u00e0 m\u1ed7i quy\u1ec3n s\u1ed5 tay l\u1ea7n l\u01b0\u1ee3t l\u00e0<\/span> ","select":["A. $12000$ \u0111\u1ed3ng, $7000$ \u0111\u1ed3ng; $3000$ \u0111\u1ed3ng.","B. $3000$ \u0111\u1ed3ng, $12000$ \u0111\u1ed3ng; $7000$ \u0111\u1ed3ng.","C. $7000$ \u0111\u1ed3ng, $3000$ \u0111\u1ed3ng; $12000$ \u0111\u1ed3ng.","D. $12000$ \u0111\u1ed3ng, $3000$ \u0111\u1ed3ng; $7000$ \u0111\u1ed3ng."],"hint":"","explain":"<span class='basic_left'>G\u1ecdi $x,y,z$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 gi\u00e1 m\u1ed9t c\u00e1i b\u00fat, m\u1ed9t quy\u1ec3n v\u1edf v\u00e0 m\u1ed9t quy\u1ec3n s\u1ed5 tay (\u0111\u01a1n v\u1ecb: \u0111\u1ed3ng)<br\/> \u0110i\u1ec1u ki\u1ec7n: $ x; y; z >0 $<br\/>B\u1ea1n th\u1ee9 nh\u1ea5t mua $2$ c\u00e1i b\u00fat v\u00e0 $8$ quy\u1ec3n v\u1edf h\u1ebft $48000$ \u0111\u1ed3ng n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: $2x+8y=48000$ (1)<br\/>B\u1ea1n th\u1ee9 hai mua $5$ quy\u1ec3n v\u1edf v\u00e0 $1$ quy\u1ec3n s\u1ed5 tay h\u1ebft $22000$ \u0111\u1ed3ng n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh $5y+z=22000$ (2)<br\/>B\u1ea1n th\u1ee9 ba mua $1$ c\u00e1i b\u00fat, $4$ quy\u1ec3n v\u1edf v\u00e0 $2$ quy\u1ec3n s\u1ed5 tay h\u1ebft $38 000$ \u0111\u1ed3ng n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh $x+4y+2z=38000$ (3)<br\/>T\u1eeb (1), (2) v\u00e0 (3), ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{align} & 2x+4y=48000\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 1 \\right) \\\\ & 5y+z=22000\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 2 \\right) \\\\ & x+4y+2z=38000\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 3 \\right) \\\\ \\end{align} \\right.$ <br\/>Bi\u1ec3u di\u1ec5n $x$ v\u00e0 $z$ theo $y$ t\u1eeb c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh (1) v\u00e0 (2), ta c\u00f3 $x=24000-4y;\\,\\,z=22000-5y$ <br\/>Thay c\u00e1c gi\u00e1 tr\u1ecb n\u00e0y v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh (3), ta c\u00f3 $\\left( 24000-4y \\right)+4y+2\\left( 22000-5y \\right)=38000$ $\\Leftrightarrow 10y=30000\\Leftrightarrow y=3000$ <br\/>T\u1eeb \u0111\u00f3 t\u00ecm \u0111\u01b0\u1ee3c $x=12000;z=7000$ <br\/>V\u1eady m\u1ed7i c\u00e1i b\u00fat, quy\u1ec3n v\u1edf, quy\u1ec3n s\u1ed5 tay gi\u00e1 l\u1ea7n l\u01b0\u1ee3t l\u00e0 $12000$ \u0111\u1ed3ng, $3000$ \u0111\u1ed3ng; $7000$ \u0111\u1ed3ng.<br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><\/span>","column":2}]}],"id_ques":398},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["12"],["30"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'>Con \u0111\u01b0\u1eddng t\u1eeb b\u1ea3n A \u0111\u1ebfn tr\u1ea1m x\u00e1 x\u00e3 g\u1ed3m m\u1ed9t \u0111o\u1ea1n l\u00ean d\u1ed1c d\u00e0i $3$ $km,$ \u0111o\u1ea1n n\u1eb1m ngang $12$ $km,$ \u0111o\u1ea1n xu\u1ed1ng d\u1ed1c $6$ $km.$ M\u1ed9t c\u00e1n b\u1ed9 \u0111i xe m\u00e1y t\u1eeb b\u1ea3n A \u0111\u1ebfn tr\u1ea1m x\u00e1 h\u1ebft $1$ gi\u1edd $7$ ph\u00fat. Sau \u0111\u00f3 c\u00e1n b\u1ed9 n\u00e0y \u0111i t\u1eeb tr\u1ea1m x\u00e1 tr\u1edf v\u1ec1 b\u1ea3n h\u1ebft $1$ gi\u1edd $16$ ph\u00fat. H\u00e3y t\u00ednh v\u1eadn t\u1ed1c c\u1ee7a xe m\u00e1y l\u00fac l\u00ean d\u1ed1c v\u00e0 l\u00fac xu\u1ed1ng d\u1ed1c bi\u1ebft r\u1eb1ng tr\u00ean \u0111o\u1ea1n \u0111\u01b0\u1eddng n\u1eb1m ngang, xe m\u00e1y \u0111i v\u1edbi v\u1eadn t\u1ed1c $18$ $km\/h$ v\u00e0 v\u1eadn t\u1ed1c khi l\u00ean d\u1ed1c, khi xu\u1ed1ng d\u1ed1c trong l\u00fac \u0111i v\u00e0 l\u00fac v\u1ec1 l\u00e0 nh\u01b0 nhau.<br\/><b>\u0110\u00e1p s\u1ed1: <\/b>V\u1eadn t\u1ed1c l\u00fac l\u00ean d\u1ed1c l\u00e0 _input_$km\/h$ v\u00e0 v\u1eadn t\u1ed1c l\u00fac xu\u1ed1ng d\u1ed1c l\u00e0 _input_$km\/h$<\/span> ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>Con \u0111\u01b0\u1eddng t\u1eeb b\u1ea3n A \u0111\u1ebfn tr\u1ea1m x\u00e1 \u0111\u01b0\u1ee3c m\u00f4 t\u1ea3 b\u1eb1ng h\u00ecnh v\u1ebd sau:<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai17/lv3/img\/D4.1a.png' \/><\/center><br\/>Ta l\u1eadp hai ph\u01b0\u01a1ng tr\u00ecnh bi\u1ec3u th\u1ecb th\u1eddi gian c\u1ee7a c\u00e1n b\u1ed9 l\u00fac \u0111i v\u00e0 l\u00fac v\u1ec1.<br\/><b> Ch\u00fa \u00fd:<\/b> L\u00fac \u0111i v\u00e0 l\u00fac v\u1ec1, \u0111o\u1ea1n l\u00ean d\u1ed1c v\u00e0 xu\u1ed1ng d\u1ed1c thay \u0111\u1ed5i nh\u01b0ng v\u1eadn t\u1ed1c khi l\u00ean d\u1ed1c, khi xu\u1ed1ng d\u1ed1c trong l\u00fac \u0111i v\u00e0 l\u00fac v\u1ec1 l\u00e0 nh\u01b0 nhau.<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>\u0110\u1ed5i $1$ gi\u1edd $7$ ph\u00fat $=1\\dfrac{7}{60}=\\dfrac{67}{60}$ (gi\u1edd); $1$ gi\u1edd $76$ ph\u00fat $=1\\dfrac{16}{60}=\\dfrac{76}{60}$ (gi\u1edd)<br\/>G\u1ecdi $x$ l\u00e0 v\u1eadn t\u1ed1c l\u00fac l\u00ean d\u1ed1c, $y$ l\u00e0 v\u1eadn t\u1ed1c l\u00fac xu\u1ed1ng d\u1ed1c. \u0110\u01a1n v\u1ecb: $km\/h.$<br\/>\u0110i\u1ec1u ki\u1ec7n: $x,y>0$<br\/>Th\u1eddi gian c\u1ee7a xe tr\u00ean \u0111o\u1ea1n n\u1eb1m ngang l\u00e0 $\\dfrac{12}{18}=\\dfrac{2}{3}$ gi\u1edd<br\/>+ \u0110i t\u1eeb b\u1ea3n A \u0111\u1ebfn tr\u1ea1m x\u00e1: Th\u1eddi gian l\u00ean d\u1ed1c c\u1ee7a xe m\u00e1y l\u00e0 $\\dfrac{3}{x}$ gi\u1edd, th\u1eddi gian xu\u1ed1ng d\u1ed1c l\u00e0 $\\dfrac{6}{y}$ gi\u1edd<br\/>Suy ra th\u1eddi gian \u0111i t\u1eeb b\u1ea3n A \u0111\u1ebfn tr\u1ea1m x\u00e1 l\u00e0 $\\dfrac{3}{x}+\\dfrac{2}{3}+\\dfrac{6}{y}$ . V\u00ec th\u1eddi gian \u0111i h\u1ebft $\\dfrac{67}{60}$ gi\u1edd n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh<br\/><\/span> $\\dfrac{3}{x}+\\dfrac{2}{3}+\\dfrac{6}{y}=\\dfrac{67}{60}$ (1)<br\/><span class='basic_left'>+ \u0110i t\u1eeb b\u1ea3n tr\u1ea1m x\u00e1 v\u1ec1 b\u1ea3n A: Th\u1eddi gian l\u00ean d\u1ed1c c\u1ee7a xe m\u00e1y l\u00e0 $\\dfrac{6}{x}$ gi\u1edd, th\u1eddi gian xu\u1ed1ng d\u1ed1c l\u00e0 $\\dfrac{3}{y}$ gi\u1edd<br\/>Suy ra th\u1eddi gian \u0111i t\u1eeb b\u1ea3n A \u0111\u1ebfn tr\u1ea1m x\u00e1 l\u00e0 $\\dfrac{6}{x}+\\dfrac{2}{3}+\\dfrac{3}{y}$. V\u00ec th\u1eddi gian \u0111i h\u1ebft $\\dfrac{76}{60}$ gi\u1edd n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh <br\/><\/span>$\\dfrac{6}{x}+\\dfrac{2}{3}+\\dfrac{3}{y}=\\dfrac{76}{60}$ (2)<br\/><span class='basic_left'>T\u1eeb (1) v\u00e0 (2), ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{aligned} & \\dfrac{3}{x}+\\dfrac{2}{3}+\\dfrac{6}{y}=\\dfrac{67}{60} \\\\ & \\dfrac{6}{x}+\\dfrac{2}{3}+\\dfrac{3}{y}=\\dfrac{76}{60} \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & \\dfrac{3}{x}+\\dfrac{6}{y}=\\dfrac{27}{60} \\\\ & \\dfrac{6}{x}+\\dfrac{3}{y}=\\dfrac{36}{60} \\\\ \\end{aligned} \\right.$ <br\/>\u0110\u1eb7t $\\dfrac{1}{x}=u;\\dfrac{1}{y}=v$ , ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{aligned} & 3u+6v=\\dfrac{27}{60} \\\\ & 6u+3v=\\dfrac{36}{60} \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & 6u+12v=\\dfrac{27}{30} \\\\ & 6u+3v=\\dfrac{36}{60} \\\\ \\end{aligned} \\right.$ <br\/>$\\Leftrightarrow \\left\\{ \\begin{aligned} & 9v=\\dfrac{3}{10} \\\\ & 6u+3v=\\dfrac{36}{60} \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & v=\\dfrac{1}{30} \\\\ & u=\\dfrac{1}{12} \\\\ \\end{aligned} \\right.$ <br\/>Suy ra $\\left\\{ \\begin{align} & x=12 \\\\ & y=30 \\\\ \\end{align} \\right.$ (th\u1ecfa m\u00e3n)<br\/>V\u1eady v\u1eadn t\u1ed1c xe m\u00e1y l\u00fac l\u00ean d\u1ed1c l\u00e0 $12$ $km\/h,$ l\u00fac xu\u1ed1ng d\u1ed1c l\u00e0 $30$ $km\/h.$<br\/><span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 l\u00e0 $12$; $30.$<\/span><\/span>"}]}],"id_ques":399},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["6,0","6"],["9,0","9"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>\u0110i\u1ec3m trung b\u00ecnh c\u1ee7a $100$ h\u1ecdc sinh trong hai l\u1edbp 9A v\u00e0 9B l\u00e0 $7,2.$ T\u00ednh \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a c\u00e1c h\u1ecdc sinh m\u1ed7i l\u1edbp bi\u1ebft r\u1eb1ng s\u1ed1 h\u1ecdc sinh l\u1edbp 9A g\u1ea5p r\u01b0\u1ee1i s\u1ed1 h\u1ecdc sinh l\u1edbp 9B v\u00e0 \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a h\u1ecdc sinh l\u1edbp 9B g\u1ea5p r\u01b0\u1ee1i \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a l\u1edbp 9A. <br\/><b> \u0110\u00e1p s\u1ed1: <\/b> \u0110i\u1ec3m trung b\u00ecnh c\u1ee7a c\u00e1c h\u1ecdc sinh l\u1edbp 9A l\u00e0 _input_<br\/> \u0110i\u1ec3m trung b\u00ecnh c\u1ee7a c\u00e1c h\u1ecdc sinh l\u1edbp 9B l\u00e0 _input_<\/span>","hint":"T\u00ednh s\u1ed1 h\u1ecdc sinh c\u1ee7a m\u1ed7i l\u1edbp","explain":"<span class='basic_left'>G\u1ecdi s\u1ed1 h\u1ecdc sinh l\u1edbp 9A l\u00e0 $a$ (h\u1ecdc sinh) v\u00e0 s\u1ed1 h\u1ecdc sinh l\u1edbp 9B l\u00e0 $b$ (h\u1ecdc sinh). \u0110i\u1ec1u ki\u1ec7n $a,$ $b>0$<br\/>V\u00ec t\u1ed5ng s\u1ed1 h\u1ecdc sinh c\u1ee7a hai l\u1edbp l\u00e0 $100$ em n\u00ean ta c\u00f3 $a+b=100$<br\/>Do s\u1ed1 h\u1ecdc sinh l\u1edbp 9A g\u1ea5p r\u01b0\u1ee1i s\u1ed1 h\u1ecdc sinh l\u1edbp 9B n\u00ean ta c\u00f3: $a=\\dfrac{3}{2}b\\Leftrightarrow \\dfrac{a}{b}=\\dfrac{3}{2}$ <br\/>$\\Leftrightarrow \\dfrac{a}{3}=\\dfrac{b}{2}=\\dfrac{a+b}{5}=\\dfrac{100}{5}=20$<br\/>Suy ra $a=3.20=60$ v\u00e0 $b=20.2=40$<br\/>V\u1eady s\u1ed1 h\u1ecdc sinh l\u1edbp 9A l\u00e0 $60$ h\u1ecdc sinh, s\u1ed1 h\u1ecdc sinh l\u1edbp 9B l\u00e0 $40$ h\u1ecdc sinh<br\/>G\u1ecdi \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a c\u00e1c h\u1ecdc sinh l\u1edbp 9A l\u00e0 $x$, \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a c\u00e1c h\u1ecdc sinh l\u1edbp 9B l\u00e0 $y$. \u0110i\u1ec1u ki\u1ec7n $x, y >0.$<br\/>V\u00ec \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a h\u1ecdc sinh l\u1edbp 9B g\u1ea5p r\u01b0\u1ee1i \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a l\u1edbp 9A n\u00ean ta c\u00f3 $y=1,5x $(1)<br\/>\u0110i\u1ec3m trung b\u00ecnh c\u1ee7a c\u00e1c h\u1ecdc sinh c\u1ea3 hai l\u1edbp l\u00e0 $7,2$ n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: $\\dfrac{60x+40y}{100}=7,2$ (2)<br\/>T\u1eeb (1) v\u00e0 (2), ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: $\\left\\{ \\begin{align} & y=1,5x \\\\ & \\dfrac{60x+40y}{100}=7,2 \\\\ \\end{align} \\right.$<br\/>$\\Leftrightarrow \\left\\{ \\begin{align} & y=1,5x \\\\ & 60x+40.1,5x=720 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & y=1,5x \\\\ & 120x=720 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & y=9 \\\\ & x=6 \\\\ \\end{align} \\right.$ (th\u1ecfa m\u00e3n)<br\/>V\u1eady \u0111i\u1ec3m trung b\u00ecnh c\u00e1c h\u1ecdc sinh l\u1edbp 9A l\u00e0 $6,0$ v\u00e0 \u0111i\u1ec3m trung b\u00ecnh c\u1ee7a c\u00e1c h\u1ecdc sinh l\u1edbp 9B l\u00e0 $9,0.$<br\/><span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 l\u00e0 $6,0$ v\u00e0 $9,0$<\/span><\/span>"}]}],"id_ques":400}],"lesson":{"save":0,"level":3}}