{"segment":[{"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":[[["72"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>Cho s\u1ed1 t\u1ef1 nhi\u00ean c\u00f3 hai ch\u1eef s\u1ed1. Bi\u1ebft ch\u1eef s\u1ed1 h\u00e0ng ch\u1ee5c l\u1edbn h\u01a1n ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb l\u00e0 $5$, n\u1ebfu vi\u1ebft ch\u1eef s\u1ed1 $0$ v\u00e0o gi\u1eefa s\u1ed1 h\u00e0ng ch\u1ee5c v\u00e0 ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb th\u00ec ta \u0111\u01b0\u1ee3c s\u1ed1 t\u1ef1 nhi\u00ean m\u1edbi l\u1edbn h\u01a1n s\u1ed1 c\u0169 $630$ \u0111\u01a1n v\u1ecb. T\u00ecm s\u1ed1 t\u1ef1 nhi\u00ean \u0111\u00f3.<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> S\u1ed1 t\u1ef1 nhi\u00ean t\u00ecm \u0111\u01b0\u1ee3c l\u00e0 _input_","hint":"G\u1ecdi ch\u1eef s\u1ed1 h\u00e0ng ch\u1ee5c l\u00e0 $a$, ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb l\u00e0 $b.$<br\/><b>Ch\u00fa \u00fd:<\/b><br\/>S\u1ed1 t\u1ef1 nhi\u00ean c\u00f3 hai ch\u1eef s\u1ed1 c\u00f3 d\u1ea1ng $\\overline{ab}=a+10b$ trong \u0111\u00f3 $a,b$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 ch\u1eef s\u1ed1 h\u00e0ng ch\u1ee5c, v\u00e0 ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb<br\/>S\u1ed1 t\u1ef1 nhi\u00ean c\u00f3 ba ch\u1eef s\u1ed1 c\u00f3 d\u1ea1ng $\\overline{abc}=100a+10b+c$, trong \u0111\u00f3 $a,b,c$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 ch\u1eef s\u1ed1 h\u00e0ng tr\u0103m, h\u00e0ng ch\u1ee5c, h\u00e0ng \u0111\u01a1n v\u1ecb.","explain":"<span class='basic_left'>G\u1ecdi $a$ l\u00e0 ch\u1eef s\u1ed1 h\u00e0ng ch\u1ee5c, $b$ l\u00e0 ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb.<br\/> \u0110i\u1ec1u ki\u1ec7n $ 0 < a \\le 9;$ $0 \\le b \\le 9$ v\u00e0 $a, b \\in \\mathbb{N}$ <br\/> Khi \u0111\u00f3 s\u1ed1 t\u1ef1 nhi\u00ean c\u1ea7n t\u00ecm l\u00e0 $\\overline{ab}$ <br\/> V\u00ec ch\u1eef s\u1ed1 h\u00e0ng ch\u1ee5c l\u1edbn h\u01a1n ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb l\u00e0 $5$ n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: $a-b=5$ (1)<br\/> Vi\u1ebft ch\u1eef s\u1ed1 $0$ v\u00e0o gi\u1eefa s\u1ed1 h\u00e0ng ch\u1ee5c v\u00e0 ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb, ta \u0111\u01b0\u1ee3c ch\u1eef s\u1ed1 m\u1edbi l\u00e0 $\\overline{a0b}$<br\/> V\u00ec s\u1ed1 m\u1edbi l\u1edbn h\u01a1n s\u1ed1 c\u0169 $630$ \u0111\u01a1n v\u1ecb n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: $\\overline{a0b}-\\overline{ab}=630$ (2)<br\/>T\u1eeb (1) v\u00e0 (2), ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{align}& a-b=5 \\\\ & \\overline{a0b}-\\overline{ab}=630 \\\\ \\end{align} \\right.$<br\/>$\\Leftrightarrow \\left\\{ \\begin{aligned} & a-b=5 \\\\ & \\left( 100a+b \\right)-\\left( 10a+b \\right)=630 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & a-b=5 \\\\ & 90a=630 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & a-b=5 \\\\ & a=7 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & b=2 \\\\ & a=7 \\\\ \\end{aligned} \\right.$(th\u1ecfa m\u00e3n)<br\/> V\u1eady s\u1ed1 t\u1ef1 nhi\u00ean c\u1ea7n t\u00ecm l\u00e0 $72$<br\/><span class='basic_pink'>V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $72$<\/span><\/span>"}]}],"id_ques":341},{"time":24,"part":[{"time":3,"title":"M\u1ed9t \u00f4 t\u00f4 \u0111i qu\u00e3ng \u0111\u01b0\u1eddng $AB$ d\u00e0i $150km$ v\u1edbi th\u1eddi gian \u0111\u00e3 \u0111\u1ecbnh. Sau khi \u0111i \u0111\u01b0\u1ee3c m\u1ed9t n\u1eeda qu\u00e3ng \u0111\u01b0\u1eddng th\u00ec \u00f4 t\u00f4 d\u1eebng l\u1ea1i $10$ ph\u00fat, do \u0111\u00f3 \u0111\u1ec3 \u0111\u1ebfn $B$ \u0111\u00fang h\u1eb9n xe ph\u1ea3i t\u0103ng v\u1eadn t\u1ed1c th\u00eam $5km\/h$ tr\u00ean qu\u00e3ng \u0111\u01b0\u1eddng c\u00f2n l\u1ea1i. T\u00ednh v\u1eadn t\u1ed1c d\u1ef1 \u0111\u1ecbnh c\u1ee7a \u00f4 t\u00f4 v\u00e0 th\u1eddi gian d\u1ef1 \u0111\u1ecbnh c\u1ee7a \u00f4 t\u00f4. ","title_trans":"H\u00e3y s\u1eafp x\u1ebfp c\u00e1c \u00fd \u0111\u1ec3 \u0111\u01b0\u1ee3c l\u1eddi gi\u1ea3i \u0111\u00fang","temp":"sequence","correct":[[[7],[3],[6],[1],[5],[2],[4]]],"list":[{"point":10,"image":"img\/1.png","left":["V\u1eady v\u1eadn t\u1ed1c d\u1ef1 \u0111\u1ecbnh c\u1ee7a \u00f4 t\u00f4 l\u00e0 $45km\/h$ v\u00e0 th\u1eddi gian d\u1ef1 \u0111\u1ecbnh c\u1ee7a \u00f4 t\u00f4 l\u00e0 $3$ gi\u1edd $20$ ph\u00fat.","K\u1ec3 t\u1eeb l\u00fac d\u1eebng $10$ ph\u00fat $(= \\dfrac{1}{6}$ gi\u1edd), \u00f4 t\u00f4 \u0111i v\u1edbi v\u1eadn t\u1ed1c $x+5$ $(km\/h)$ v\u00e0 th\u1eddi gian k\u1ec3 t\u1eeb l\u00fac d\u1eebng cho t\u1edbi khi \u0111\u1ebfn $B$ l\u00e0 $\\dfrac{y}{2}+\\dfrac{1}{6}$ .","Gi\u1ea3i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c $x=45$ v\u00e0 $y=\\dfrac{10}{3}$ (th\u1ecfa m\u00e3n)","G\u1ecdi v\u1eadn t\u1ed1c d\u1ef1 \u0111\u1ecbnh l\u00e0 $x$ $(km\/h)$, $x > 0$ v\u00e0 th\u1eddi gian d\u1ef1 \u0111\u1ecbnh l\u00e0 $y$ ( gi\u1edd), $y > 0$","T\u1eeb (1) v\u00e0 (2), ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: $\\left\\{ \\begin{align}& xy=150 \\\\ & \\left( x+5 \\right)\\left( \\dfrac{y}{2}+\\dfrac{1}{6} \\right)=75 \\\\ \\end{align} \\right.$ ","Do \u0111\u1ed9 d\u00e0i qu\u00e3ng \u0111\u01b0\u1eddng $AB$ l\u00e0 $150$ $km$ n\u00ean ta c\u00f3 $xy=150$ (1)"," \u0110\u1ed9 d\u00e0i n\u1eeda qu\u00e3ng \u0111\u01b0\u1eddng sau l\u00e0 $75$ $km$ n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:$\\left( x+5 \\right)\\left( \\dfrac{y}{2}+\\dfrac{1}{6} \\right)=75$ (2)"],"top":100,"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>\u0110\u1ed5i $10$ ph\u00fat $= \\dfrac{1}{6}$ gi\u1edd<br\/> B\u1ea3ng ph\u00e2n t\u00edch chuy\u1ec3n \u0111\u1ed9ng c\u1ee7a \u00f4 t\u00f4:<br\/><table><tr><th>C\u00e1c \u0111\u1ea1i l\u01b0\u1ee3ng<br><\/th><th>V\u1eadn t\u1ed1c $(km\/h)$<br><\/th><th>Qu\u00e3ng \u0111\u01b0\u1eddng $(km)$<br><\/th><th>Th\u1eddi gian (gi\u1edd)<br><\/th><\/tr><tr><th>D\u1ef1 \u0111\u1ecbnh<br><\/th><td>$x$<\/td><td>$150$<\/td><td>$y$<\/td><\/tr><tr><th>Th\u1ef1c t\u1ebf<br\/> (x\u00e9t n\u1eeda qu\u00e3ng \u0111\u01b0\u1eddng sau)<br><\/th><td>$x+5$<\/td><td>$75$<\/td><td>$\\dfrac{y}{2}+\\dfrac{1}{6}$<\/td><\/tr><\/table><br\/>- L\u1eadp ph\u01b0\u01a1ng tr\u00ecnh bi\u1ec3u th\u1ecb \u0111\u1ed9 d\u00e0i qu\u00e3ng \u0111\u01b0\u1eddng $AB$ v\u00e0 ph\u01b0\u01a1ng tr\u00ecnh bi\u1ec3u th\u1ecb chuy\u1ec3n \u0111\u1ed9ng c\u1ee7a \u00f4 t\u00f4 trong n\u1eeda qu\u00e3ng \u0111\u01b0\u1eddng sau.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>\u0110\u1ed5i $10$ ph\u00fat $= \\dfrac{1}{6}$ gi\u1edd<br\/>G\u1ecdi v\u1eadn t\u1ed1c d\u1ef1 \u0111\u1ecbnh l\u00e0 $x$ $(km\/h)$, $x > 0$ v\u00e0 th\u1eddi gian d\u1ef1 \u0111\u1ecbnh l\u00e0 $y$ ( gi\u1edd), $y > 0$<br\/>Do \u0111\u1ed9 d\u00e0i qu\u00e3ng \u0111\u01b0\u1eddng $AB$ l\u00e0 $150$ $km$ n\u00ean ta c\u00f3 $xy=150$ (1)<br\/>K\u1ec3 t\u1eeb l\u00fac d\u1eebng $10$ ph\u00fat, \u00f4 t\u00f4 \u0111i v\u1edbi v\u1eadn t\u1ed1c $x+5$ $(km\/h)$ v\u00e0 th\u1eddi gian k\u1ec3 t\u1eeb l\u00fac d\u1eebng cho t\u1edbi khi \u0111\u1ebfn $B$ l\u00e0 $\\dfrac{y}{2}+\\dfrac{1}{6}.$ <br\/> \u0110\u1ed9 d\u00e0i n\u1eeda qu\u00e3ng \u0111\u01b0\u1eddng sau l\u00e0 $75$ $km$ n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:$\\left( x+5 \\right)\\left( \\dfrac{y}{2}+\\dfrac{1}{6} \\right)=75$ (2)<br\/>T\u1eeb (1) v\u00e0 (2), ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: $\\left\\{ \\begin{align}& xy=150 \\\\ & \\left( x+5 \\right)\\left( \\dfrac{y}{2}+\\dfrac{1}{6} \\right)=75 \\\\ \\end{align} \\right.$<br\/> $\\Leftrightarrow \\left\\{ \\begin{aligned} & xy=150 \\\\ & \\dfrac{xy}{2}-\\dfrac{x}{6}+\\dfrac{5y}{2}-\\dfrac{5}{6}=75 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & xy=150 \\\\ & -\\dfrac{x}{6}+\\dfrac{5y}{2}=\\dfrac{5}{6} \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & xy=150 \\\\ & x=15y-5 \\\\ \\end{aligned} \\right.$ <br\/>$\\Leftrightarrow \\left\\{ \\begin{aligned} & \\left( 15y-5 \\right)y=150 \\\\ & x=15y-5 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & 15{{y}^{2}}-5y=150 \\\\ & x=15y-5 \\\\ \\end{aligned} \\right.$<br\/>$\\left( 3 \\right)\\Leftrightarrow \\left( 3y-10 \\right)\\left( y+3 \\right)=0\\Leftrightarrow \\left[ \\begin{aligned} & y=\\dfrac{10}{3} \\\\ & y=-3 \\\\ \\end{aligned} \\right.\\Leftrightarrow y=\\dfrac{10}{3}$ <br\/>Suy ra $x=45$<br\/>V\u1eady v\u1eadn t\u1ed1c d\u1ef1 \u0111\u1ecbnh c\u1ee7a \u00f4 t\u00f4 l\u00e0 $45km\/h$ v\u00e0 th\u1eddi gian d\u1ef1 \u0111\u1ecbnh c\u1ee7a \u00f4 t\u00f4 l\u00e0 $3$ gi\u1edd $20$ ph\u00fat.<\/span><\/span>"}]}],"id_ques":342},{"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":[[["9"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"<span class='basic_left'>M\u1ed9t ph\u00e2n x\u01b0\u1edfng s\u1ea3n xu\u1ea5t \u0111\u1ec1 ra k\u1ebf ho\u1ea1ch s\u1ea3n xu\u1ea5t $180$ t\u1ea5n s\u1ea3n ph\u1ea9m. Khi th\u1ef1c hi\u1ec7n m\u1ed7i ng\u00e0y \u0111\u00e3 v\u01b0\u1ee3t m\u1ee9c $1$ t\u1ea5n so v\u1edbi k\u1ebf ho\u1ea1ch n\u00ean ch\u1eb3ng nh\u1eefng xong s\u1edbm $1$ ng\u00e0y m\u00e0 c\u00f2n l\u00e0m th\u00eam \u0111\u01b0\u1ee3c $10$ t\u1ea5n s\u1ea3n ph\u1ea7m. T\u00ednh n\u0103ng su\u1ea5t d\u1ef1 ki\u1ebfn theo k\u1ebf ho\u1ea1ch. <br\/><b>\u0110\u00e1p s\u1ed1:<\/b> N\u0103ng su\u1ea5t d\u1ef1 ki\u1ebfn theo k\u1ebf ho\u1ea1ch l\u00e0 _input_ (t\u1ea5n\/ ng\u00e0y)<\/span>","hint":"G\u1ecdi \u1ea9n l\u00e0 th\u1eddi gian x\u01b0\u1edfng ph\u1ea3i ho\u00e0n th\u00e0nh theo k\u1ebf ho\u1ea1ch v\u00e0 th\u1eddi gian x\u01b0\u1edfng ho\u00e0n th\u00e0nh theo th\u1ef1c t\u1ebf","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span> Ph\u00e2n t\u00edch b\u00e0i to\u00e1n<br\/><table><tr><th>C\u00e1c \u0111\u1ea1i l\u01b0\u1ee3ng<br><\/th><th>S\u1ed1 s\u1ea3n ph\u1ea9m<br><\/th><th>Th\u1eddi gian (ng\u00e0y)<br><\/th><th>N\u0103ng su\u1ea5t (s\u1ea3n ph\u1ea9m\/ng\u00e0y)<br><\/th><\/tr><tr><th>D\u1ef1 ki\u1ebfn<br><\/th><td>$180$<\/td><td>$x$<\/td><td>$\\dfrac{180}{x}$<\/td><\/tr><tr><th>Th\u1ef1c t\u1ebf<br><\/th><td>$190$<\/td><td>$y$<\/td><td>$\\dfrac{190}{y}$<\/td><\/tr><\/table><br\/>T\u1eeb gi\u1ea3 thi\u1ebft b\u00e0i to\u00e1n, ta \u0111i l\u1eadp c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ec3 t\u00ecm $x$ v\u00e0 $y$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>G\u1ecdi th\u1eddi gian x\u01b0\u1edfng ph\u1ea3i ho\u00e0n th\u00e0nh theo k\u1ebf ho\u1ea1ch l\u00e0 $x$ ( ng\u00e0y ); th\u1eddi gian x\u01b0\u1edfng ho\u00e0n th\u00e0nh theo th\u1ef1c t\u1ebf l\u00e0 $y$ (ng\u00e0y ).<br\/> \u0110i\u1ec1u ki\u1ec7n: $x, y >0.$<br\/>V\u00ec th\u1ef1c t\u1ebf xong s\u1edbm $1$ ng\u00e0y theo k\u1ebf ho\u1ea1ch n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: $x-y=1$ (1)<br\/>Theo k\u1ebf ho\u1ea1ch, $1$ ng\u00e0y x\u01b0\u1edfng s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c $\\dfrac{180}{x}$ t\u1ea5n s\u1ea3n ph\u1ea9m <br\/>Th\u1ef1c t\u1ebf, x\u01b0\u1edfng s\u1ea3n xu\u1ea5t v\u01b0\u1ee3t m\u1ee9c $10$ s\u1ea3n ph\u1ea9m, t\u1ee9c s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c $190$ s\u1ea3n ph\u1ea9m n\u00ean n\u0103ng su\u1ea5t trong $1$ ng\u00e0y l\u00e0 $\\dfrac{190}{y}$ t\u1ea5n s\u1ea3n ph\u1ea9m.<br\/> V\u00ec th\u1ef1c t\u1ebf m\u1ed7i ng\u00e0y v\u01b0\u1ee3t m\u1ee9c $1$ t\u1ea5n so v\u1edbi k\u1ebf ho\u1ea1ch n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/><\/span> $\\dfrac{180}{x}+1=\\dfrac{190}{y}$ (2) <br\/><span class='basic_left'>T\u1eeb (1) v\u00e0 (2), ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh <\/span> $\\left\\{ \\begin{aligned}& x-y=1 \\, (1) \\\\ & \\dfrac{180}{x}+1=\\dfrac{190}{y} \\, (2) \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x=1+y \\, (1') \\\\ & \\dfrac{180}{x}+1=\\dfrac{190}{y} \\,(2) \\\\ \\end{aligned} \\right.$<br\/><span class='basic_left'>Thay (1\u2019) v\u00e0o (2), ta \u0111\u01b0\u1ee3c: <br\/>$\\left( 2 \\right)\\Leftrightarrow \\dfrac{180}{y+1}+1=\\dfrac{190}{y}\\Leftrightarrow 180y+y\\left( y+1 \\right)=190\\left( y+1 \\right)$ <br\/>$\\Leftrightarrow {{y}^{2}}-9y-190=0\\Leftrightarrow \\left( y-19 \\right)\\left( y+10 \\right)=0$<br\/>$\\Leftrightarrow \\left[ \\begin{align} & y=19 \\\\ & y=-10 \\\\ \\end{align} \\right.\\Leftrightarrow y=19$<br\/>Suy ra $x = 1 + y=20$ (th\u1ecfa m\u00e3n)<br\/>V\u1eady n\u0103ng su\u1ea5t d\u1ef1 ki\u1ebfn theo k\u1ebf ho\u1ea1ch l\u00e0 $\\dfrac{180}{20}=9$ t\u1ea5n s\u1ea3n ph\u1ea9m m\u1ed7i ng\u00e0y.<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $9.$ <\/span><br\/><span class='basic_green'><b>L\u01b0u \u00fd:<\/b><\/span> C\u00f3 th\u1ec3 gi\u1ea3i b\u00e0i to\u00e1n b\u1eb1ng c\u00e1ch gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: <br\/>G\u1ecdi th\u1eddi gian x\u01b0\u1edfng ph\u1ea3i ho\u00e0n th\u00e0nh theo k\u1ebf ho\u1ea1ch l\u00e0 $x$ ( ng\u00e0y ). Khi \u0111\u00f3 th\u1eddi gian x\u01b0\u1edfng ho\u00e0n th\u00e0nh theo th\u1ef1c t\u1ebf l\u00e0 $x-1$ ( ng\u00e0y ).<\/span>"}]}],"id_ques":343},{"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'>Hai b\u1ebfn s\u00f4ng $A$ v\u00e0 $B$ c\u00e1ch nhau $40$ $km.$ M\u1ed9t ca n\u00f4 xu\u00f4i d\u00f2ng t\u1eeb $A$ \u0111\u1ebfn $B$ r\u1ed3i quay tr\u1edf l\u1ea1i $A$ v\u1edbi v\u1eadn t\u1ed1c ri\u00eang kh\u00f4ng \u0111\u1ed5i h\u1ebft t\u1ea5t c\u1ea3 $2$ gi\u1edd $15$ ph\u00fat. Khi ca n\u00f4 kh\u1edfi h\u00e0nh t\u1eeb $A$ th\u00ec c\u00f9ng l\u00fac \u0111\u00f3, m\u1ed9t kh\u00fac g\u1ed7 c\u0169ng tr\u00f4i t\u1ef1 do t\u1eeb $A$ theo d\u00f2ng n\u01b0\u1edbc v\u00e0 g\u1eb7p ca n\u00f4 tr\u00ean \u0111\u01b0\u1eddng v\u1ec1 t\u1ea1i m\u1ed9t \u0111i\u1ec3m c\u00e1ch $A$ l\u00e0 $8$ $km.$ <br\/><b> C\u00e2u 1: <\/b> G\u1ecdi v\u1eadn t\u1ed1c ri\u00eang c\u1ee7a ca n\u00f4 l\u00e0 $x$ $(km\/h),$ v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc l\u00e0 $y$ $(km\/h)$ th\u00ec h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh l\u1eadp \u0111\u01b0\u1ee3c l\u00e0 <\/span>","select":["A. $\\left\\{ \\begin{matrix} \\dfrac{40}{x+y}+\\dfrac{40}{x-y}=\\dfrac{9}{4} \\\\ \\dfrac{40}{x+y}+\\dfrac{32}{x-y}=\\dfrac{1}{y} \\\\\\end{matrix} \\right.$","B. $\\left\\{ \\begin{matrix} \\dfrac{40}{x+y}+\\dfrac{40}{x-y}=\\dfrac{9}{4} \\\\ \\dfrac{5}{x+y}+\\dfrac{4}{x-y}=\\dfrac{8}{y} \\\\\\end{matrix} \\right.$","C. $\\left\\{ \\begin{matrix} \\dfrac{40}{x+y}+\\dfrac{40}{x-y}=\\dfrac{9}{4} \\\\ \\dfrac{5}{x+y}+\\dfrac{4}{x-y}=\\dfrac{1}{y} \\\\\\end{matrix} \\right.$","D. $\\left\\{ \\begin{matrix} \\dfrac{40}{x+y}+\\dfrac{40}{x-y}=\\dfrac{9}{4} \\\\ \\dfrac{4}{x+y}+\\dfrac{5}{x-y}=\\dfrac{1}{y} \\\\\\end{matrix} \\right.$"],"hint":"V\u1eadn t\u1ed1c xu\u00f4i d\u00f2ng $=$ v\u1eadn t\u1ed1c th\u1ef1c $+$ v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc;<br\/> V\u1eadn t\u1ed1c ng\u01b0\u1ee3c d\u00f2ng $=$ v\u1eadn t\u1ed1c th\u1ef1c $-$ v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc","explain":"<span class='basic_left'> \u0110\u1ed5i: $2$ gi\u1edd $15$ ph\u00fat = $\\dfrac {9}{4}h$ <br\/> G\u1ecdi v\u1eadn t\u1ed1c ri\u00eang c\u1ee7a ca n\u00f4 l\u00e0 $x$ $(km\/h),$ v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc l\u00e0 $y$ $(km\/h)$. \u0110i\u1ec1u ki\u1ec7n $x,y>0$<br\/> $\\Rightarrow$ V\u1eadn t\u1ed1c ca n\u00f4 khi xu\u00f4i d\u00f2ng l\u00e0 $x + y$ $(km\/h),$ v\u1eadn t\u1ed1c ca n\u00f4 khi ng\u01b0\u1ee3c d\u00f2ng l\u00e0 $x - y$ $(km\/h).$ <br\/> V\u00ec th\u1eddi gian c\u1ea3 \u0111i l\u1eabn v\u1ec1 c\u1ee7a ca n\u00f4 l\u00e0 $\\dfrac {9}{4}h$ n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: $\\dfrac{ 40}{x + y} + \\dfrac{ 40}{x - y} = \\dfrac{ 9}{4}$ (1) <br\/>Th\u1eddi gian ca n\u00f4 \u0111i xu\u00f4i $40$ $km$ v\u00e0 \u0111i ng\u01b0\u1ee3c $32$ $km$ l\u00e0: $\\dfrac{ 40}{x + y} + \\dfrac{ 32}{x - y}$ $(h).$<br\/>Kh\u00fac g\u1ed7 tr\u00f4i \u0111\u01b0\u1ee3c $8$ $km$ v\u00e0 c\u00f3 v\u1eadn t\u1ed1c b\u1eb1ng v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc n\u00ean th\u1eddi gian kh\u00fac g\u1ed7 tr\u00f4i l\u00e0 $\\dfrac{ 8}{y}$ $(h).$<br\/> Hai th\u1eddi gian b\u1eb1ng nhau n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: <br\/> $\\dfrac{ 40}{x + y} + \\dfrac{ 32}{x - y} = \\dfrac{ 8}{y}$ $\\Leftrightarrow \\dfrac{ 5}{x + y} + \\dfrac{ 4}{x - y} = \\dfrac{ 1}{y}$ (2)<br\/>T\u1eeb (1) v\u00e0 (2) ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh:<\/span><br\/> $\\left\\{ \\begin{matrix} \\dfrac{40}{x+y}+\\dfrac{40}{x-y}=\\dfrac{9}{4} \\\\ \\dfrac{5}{x+y}+\\dfrac{4}{x-y}=\\dfrac{1}{y} \\\\\\end{matrix} \\right.$ <br\/><span class='basic_left'><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":2}]}],"id_ques":344},{"time":24,"part":[{"title":"\u0110i\u1ec1n c\u00e1c s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["36"],["4"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>Hai b\u1ebfn s\u00f4ng $A$ v\u00e0 $B$ c\u00e1ch nhau $40$ $km.$ M\u1ed9t ca n\u00f4 xu\u00f4i d\u00f2ng t\u1eeb $A$ \u0111\u1ebfn $B$ r\u1ed3i quay tr\u1edf l\u1ea1i $A$ v\u1edbi v\u1eadn t\u1ed1c ri\u00eang kh\u00f4ng \u0111\u1ed5i h\u1ebft t\u1ea5t c\u1ea3 $2$ gi\u1edd $15$ ph\u00fat. Khi ca n\u00f4 kh\u1edfi h\u00e0nh t\u1eeb $A$ th\u00ec c\u00f9ng l\u00fac \u0111\u00f3, m\u1ed9t kh\u00fac g\u1ed7 c\u0169ng tr\u00f4i t\u1ef1 do t\u1eeb $A$ theo d\u00f2ng n\u01b0\u1edbc v\u00e0 g\u1eb7p ca n\u00f4 tr\u00ean \u0111\u01b0\u1eddng v\u1ec1 t\u1ea1i m\u1ed9t \u0111i\u1ec3m c\u00e1ch $A$ $8$ $km.$ <br\/><b> C\u00e2u 2: <\/b>T\u00ednh v\u1eadn t\u1ed1c ri\u00eang c\u1ee7a ca n\u00f4 v\u00e0 v\u1eadn t\u1ed1c c\u1ee7a d\u00f2ng n\u01b0\u1edbc<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> V\u1eadn t\u1ed1c ri\u00eang c\u1ee7a ca n\u00f4 l\u00e0 _input_$(km\/h)$ v\u00e0 v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc l\u00e0 _input_$(km\/h)$ ","explain":"<span class='basic_left'> G\u1ecdi v\u1eadn t\u1ed1c ri\u00eang c\u1ee7a ca n\u00f4 l\u00e0 $x$ $(km\/h),$ v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc l\u00e0 $y$ $(km\/h)$. \u0110i\u1ec1u ki\u1ec7n $x,y>0$.<br\/> Theo c\u00e2u 1, ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: $\\left\\{ \\begin{matrix} \\dfrac{40}{x+y}+\\dfrac{40}{x-y}=\\dfrac{9}{4} (1) \\\\ \\dfrac{5}{x+y}+\\dfrac{4}{x-y}=\\dfrac{1}{y} (2) \\\\\\end{matrix} \\right.$<br\/>$(2) \\Leftrightarrow y\\left[5(x- y) + 4(x + y)\\right] = (x- y)(x+y) \\Leftrightarrow y(9x-y)=x^2 - y^2$$\\Leftrightarrow x^2 = 9xy$<br\/>$\\Leftrightarrow \\left[ \\begin{align} & x=9y \\\\ & x=0\\, (\\text{lo\u1ea1i v\u00ec}\\, x>0) \\\\ \\end{align} \\right.\\Rightarrow x=9y$<br\/>Thay $x=9y$ v\u00e0o (1), ta c\u00f3 $\\dfrac{40}{9y+y}+\\dfrac{40}{9y-y}=\\dfrac{9}{4}\\Leftrightarrow \\dfrac{9}{y}=\\dfrac{9}{4}\\Leftrightarrow y=4$ (th\u1ecfa m\u00e3n)<br\/>Suy ra $x=9y=9.4=36$ (th\u1ecfa m\u00e3n)<br\/>V\u1eady v\u1eadn t\u1ed1c ri\u00eang c\u1ee7a ca n\u00f4 l\u00e0 $36$ $km\/h$, v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc l\u00e0 $4$ $km\/h.$<br\/><span class='basic_pink'>V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o hai \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $36$ v\u00e0 $4$ <\/span><\/span>"}]}],"id_ques":345},{"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":[[["100"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<span class='basic_left'>M\u1ed9t nh\u00f3m th\u1ee3 \u0111\u1eb7t k\u1ebf ho\u1ea1ch s\u1ea3n xu\u1ea5t $3000$ s\u1ea3n ph\u1ea9m. Trong $8$ ng\u00e0y \u0111\u1ea7u h\u1ecd th\u1ef1c hi\u1ec7n \u0111\u00fang k\u1ebf ho\u1ea1ch \u0111\u1ec1 ra, nh\u1eefng ng\u00e0y c\u00f2n l\u1ea1i h\u1ecd v\u01b0\u1ee3t k\u1ebf ho\u1ea1ch m\u1ed7i ng\u00e0y $10$ s\u1ea3n ph\u1ea9m, n\u00ean \u0111\u00e3 ho\u00e0n th\u00e0nh s\u1edbm h\u01a1n k\u1ebf ho\u1ea1ch $2$ ng\u00e0y. H\u1ecfi theo k\u1ebf ho\u1ea1ch m\u1ed7i ng\u00e0y nh\u00f3m th\u1ee3 s\u1ea3n xu\u1ea5t bao nhi\u00eau s\u1ea3n ph\u1ea9m.<br\/><b> \u0110\u00e1p s\u1ed1: <\/b>S\u1ed1 s\u1ea3n ph\u1ea9m theo k\u1ebf ho\u1ea1ch l\u00e0m trong m\u1ed9t ng\u00e0y l\u00e0 _input_ (s\u1ea3n ph\u1ea9m)<\/span>","hint":"G\u1ecdi \u1ea9n l\u00e0 s\u1ed1 s\u1ea3n ph\u1ea9m theo k\u1ebf ho\u1ea1ch l\u00e0m trong m\u1ed9t ng\u00e0y v\u00e0 th\u1eddi gian d\u1ef1 \u0111\u1ecbnh \u0111\u1ec3 ho\u00e0n th\u00e0nh s\u1ea3n xu\u1ea5t.","explain":"<span class='basic_left'>G\u1ecdi s\u1ed1 s\u1ea3n ph\u1ea9m theo k\u1ebf ho\u1ea1ch l\u00e0m trong m\u1ed9t ng\u00e0y l\u00e0 $x$ ( s\u1ea3n ph\u1ea9m ).<br\/>G\u1ecdi th\u1eddi gian d\u1ef1 \u0111\u1ecbnh l\u00e0 $y$ ( ng\u00e0y ). \u0110i\u1ec1u ki\u1ec7n $x, y>0$<br\/>Theo k\u1ebf ho\u1ea1ch nh\u00f3m th\u1ee3 s\u1ea3n xu\u1ea5t $300$ s\u1ea3n ph\u1ea9m n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: $xy=3000$ (1) <br\/>Trong $8$ ng\u00e0y \u0111\u1ea7u, nh\u00f3m th\u1ee3 s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c l\u00e0 $8x$ s\u1ea3n ph\u1ea9m <br\/> Nh\u1eefng ng\u00e0y c\u00f2n l\u1ea1i, h\u1ecd v\u01b0\u1ee3t k\u1ebf ho\u1ea1ch m\u1ed7i ng\u00e0y $10$ s\u1ea3n ph\u1ea9m n\u00ean trong $1$ ng\u00e0y, h\u1ecd s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c $x+10$ s\u1ea3n ph\u1ea9m <br\/> Do ho\u00e0n th\u00e0nh s\u1edbm h\u01a1n k\u1ebf ho\u1ea1ch $2$ ng\u00e0y n\u00ean th\u1eddi gian th\u1ef1c t\u1ebf (t\u00ednh sau $8$ ng\u00e0y \u0111\u1ea7u) l\u00e0 $y - 8 - 2 =y-10 $ (ng\u00e0y) <br\/> V\u00ec s\u1ed1 s\u1ea3n ph\u1ea9m c\u1ea7n s\u1ea3n xu\u1ea5t kh\u00f4ng thay \u0111\u1ed5i n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: <br\/><\/span> $8x+\\left( x+10 \\right)\\left( y-10 \\right)=3000$ (2)<br\/><span class='basic_left'>T\u1eeb (1) v\u00e0 (2), ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{align}& xy=3000 \\\\ & 8x+\\left( x+10 \\right)\\left( y-10 \\right)=3000 \\\\ \\end{align} \\right.$ <br\/>$\\Leftrightarrow \\left\\{ \\begin{aligned} & xy=3000 \\\\ & 10y-2x+xy=3100 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & xy=3000 \\\\ & 10y-2x=100 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & xy=3000\\, \\\\ & x=5y-50 \\\\ \\end{aligned} \\right.$<br\/>$\\Leftrightarrow \\left\\{ \\begin{aligned} & \\left( 5y-50 \\right)y=3000 \\\\ & x=5y-50 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & {{y}^{2}}-10y-600=0 \\\\ & x=5y-50 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & \\left( y+20 \\right)\\left( y-30 \\right)=0 \\\\ & x=5y-50 \\\\ \\end{aligned} \\right.$<br\/>$\\Leftrightarrow \\left\\{ \\begin{align} & y=30 \\,\\text{ho\u1eb7c}\\, y=-20 \\\\ & x=5y-50 \\\\ \\end{align} \\right.$$\\Leftrightarrow \\left\\{ \\begin{align} & y=30 \\,(do\\, y>0 ) \\\\ & x=100 \\\\ \\end{align} \\right.$<br\/>V\u1eady s\u1ed1 s\u1ea3n ph\u1ea9m theo k\u1ebf ho\u1ea1ch l\u00e0m trong m\u1ed9t ng\u00e0y l\u00e0 $100$ (s\u1ea3n ph\u1ea9m )<br\/><span class='basic_pink'> Do \u0111\u00f3 \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0 $100$.<\/span><\/span>."}]}],"id_ques":346},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["5"],["12"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'>T\u00ecm \u0111\u1ed9 d\u00e0i hai c\u1ea1nh g\u00f3c vu\u00f4ng c\u1ee7a m\u1ed9t tam gi\u00e1c vu\u00f4ng, bi\u1ebft r\u1eb1ng n\u1ebfu t\u0103ng m\u1ed7i c\u1ea1nh l\u00ean $2$ $cm$ th\u00ec di\u1ec7n t\u00edch tam gi\u00e1c \u0111\u00f3 s\u1ebd t\u0103ng th\u00eam $19$ $ cm^2$. N\u1ebfu m\u1ed9t c\u1ea1nh gi\u1ea3m \u0111i $1$ $cm$ v\u00e0 c\u1ea1nh kia gi\u1ea3m \u0111i $3$ $cm$ th\u00ec di\u1ec7n t\u00edch tam gi\u00e1c gi\u1ea3m \u0111i $12$ $cm^2$. <br\/><b> \u0110\u00e1p s\u1ed1: <\/b> \u0110\u1ed9 d\u00e0i hai c\u1ea1nh g\u00f3c vu\u00f4ng l\u00e0 _input_ $(cm)$ v\u00e0 _input_ $(cm)$<\/span>","hint":"Di\u1ec7n t\u00edch tam gi\u00e1c vu\u00f4ng b\u1eb1ng n\u1eeda t\u00edch hai c\u1ea1nh g\u00f3c vu\u00f4ng","explain":"<span class='basic_left'>G\u1ecdi \u0111\u1ed9 d\u00e0i c\u00e1c c\u1ea1nh g\u00f3c vu\u00f4ng l\u00e0 $x$ $(cm)$ , $y$ $(cm)$. <br\/>\u0110i\u1ec1u ki\u1ec7n $x, y >0$<br\/>Khi \u0111\u00f3 di\u1ec7n t\u00edch tam gi\u00e1c l\u00e0 $\\dfrac{xy}{2}$ $(cm^2)$<br\/>+ T\u0103ng m\u1ed7i c\u1ea1nh l\u00ean $2$ $cm$ th\u00ec di\u1ec7n t\u00edch tam gi\u00e1c l\u00e0 $\\dfrac{\\left( x+2 \\right)\\left( y+2 \\right)}{2}$ $(cm^2).$<br\/> Khi \u0111\u00f3 di\u1ec7n t\u00edch ban \u0111\u1ea7u t\u0103ng th\u00eam $19$ $cm^2$ n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: <br\/> <\/span>$\\dfrac{\\left( x+2 \\right)\\left( y+2 \\right)}{2}=\\dfrac{xy}{2}+19$(1) <br\/><span class='basic_left'>+ M\u1ed9t c\u1ea1nh gi\u1ea3m \u0111i $1$ $cm$ v\u00e0 c\u1ea1nh kia gi\u1ea3m \u0111i $3$ $cm$ th\u00ec di\u1ec7n t\u00edch tam gi\u00e1c l\u00e0 $\\dfrac{\\left( x-1 \\right)\\left( y-3 \\right)}{2}$ $(cm^2)$. <br\/>Do di\u1ec7n t\u00edch ban \u0111\u1ea7u gi\u1ea3m \u0111i $12$ $cm^2 $n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: <br\/><\/span> $\\dfrac{\\left( x-1 \\right)\\left( y-3 \\right)}{2}=\\dfrac{xy}{2}-12$(2)<br\/><span class='basic_left'>T\u1eeb (1) v\u00e0 (2), ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh <br\/> $\\left\\{ \\begin{align} & \\dfrac{\\left( x+2 \\right)\\left( y+2 \\right)}{2}=\\dfrac{xy}{2}+19 \\\\ & \\dfrac{\\left( x-1 \\right)\\left( y-3 \\right)}{2}=\\dfrac{xy}{2}-12 \\\\ \\end{align} \\right. \\Leftrightarrow \\left\\{ \\begin{aligned} & xy+2x+2y+4=xy+38 \\\\ & xy-3x-y+3=xy-24 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & 2x+2y=34 \\\\ & 3x+y=27 \\\\ \\end{aligned} \\right.$ <br\/>$\\Leftrightarrow \\left\\{ \\begin{aligned} & x+y=17 \\\\ & 3x+y=27 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & 2x=10 \\\\ & x+y=17 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x=5 \\\\ & y=12 \\\\ \\end{aligned} \\right.$(th\u1ecfa m\u00e3n)<br\/>V\u1eady \u0111\u1ed9 d\u00e0i hai c\u1ea1nh g\u00f3c vu\u00f4ng l\u00e0 $5$ $cm$ v\u00e0 $12$ $cm.$<br\/><span class='basic_pink'> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $5;12$<\/span><\/span>"}]}],"id_ques":347},{"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\u1ea3 s\u1eed c\u00f3 m\u1ed9t c\u00e1nh \u0111\u1ed3ng c\u1ecf d\u00e0y nh\u01b0 nhau, m\u1ecdc cao \u0111\u1ec1u nh\u01b0 nhau tr\u00ean to\u00e0n b\u1ed9 c\u00e1nh \u0111\u1ed3ng trong su\u1ed1t th\u1eddi gian b\u00f2 \u0103n c\u1ecf tr\u00ean c\u00e1nh \u0111\u1ed3ng \u1ea5y. Bi\u1ebft r\u1eb1ng $9$ con b\u00f2 \u0103n h\u1ebft c\u1ecf tr\u00ean c\u00e1nh \u0111\u1ed3ng trong $2$ tu\u1ea7n, $6$ con b\u00f2 \u0103n h\u1ebft c\u1ecf tr\u00ean c\u00e1nh \u0111\u1ed3ng trong $4$ tu\u1ea7n. Coi m\u1ed7i con b\u00f2 \u0103n s\u1ed1 c\u1ecf nh\u01b0 nhau. S\u1ed1 con b\u00f2 \u0103n h\u1ebft c\u1ecf trong $6$ tu\u1ea7n l\u00e0: <\/span>","select":["A. $3$ ","B. $4$","C. $5$","D. $6$"],"hint":"Coi kh\u1ed1i l\u01b0\u1ee3ng c\u1ecf c\u00f3 s\u1eb5n tr\u00ean c\u00e1nh \u0111\u1ed3ng khi b\u00f2 \u0103n c\u1ecf l\u00e0 $1$ (\u0111\u01a1n v\u1ecb kh\u1ed1i l\u01b0\u1ee3ng quy \u01b0\u1edbc). Ch\u00fa \u00fd: Trong su\u1ed1t th\u1eddi gian b\u00f2 \u0103n c\u1ecf, c\u1ecf v\u1eabn m\u1ecdc \u0111\u1ec1u tr\u00ean c\u00e1nh \u0111\u1ed3ng.<br\/> Gi\u1ea3i b\u00e0i to\u00e1n b\u1eb1ng c\u00e1ch l\u1eadp h\u1ec7 v\u1edbi 2 \u1ea9n l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng c\u1ecf m\u1ecdc th\u00eam tr\u00ean c\u00e1nh \u0111\u1ed3ng trong m\u1ed9t tu\u1ea7n v\u00e0 s\u1ed1 b\u00f2 ph\u1ea3i t\u00ecm.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/>B\u01b0\u1edbc 1: L\u1eadp h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh<br\/>+ G\u1ecdi \u1ea9n l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng c\u1ecf m\u1ecdc th\u00eam tr\u00ean c\u00e1nh \u0111\u1ed3ng trong m\u1ed9t tu\u1ea7n v\u00e0 s\u1ed1 b\u00f2 ph\u1ea3i t\u00ecm.<br\/>+ X\u00e1c \u0111\u1ecbnh l\u01b0\u1ee3ng c\u1ecf m\u1ed7i con b\u00f2 \u0103n trong $1$ tu\u1ea7n trong c\u00e1c tr\u01b0\u1eddng h\u1ee3p: 9 con \u0103n trong $2$ tu\u1ea7n, $6$ con \u0103n trong $4$ tu\u1ea7n v\u00e0 $x$ con \u0103n trong $6$ tu\u1ea7n <br\/>+ L\u1eadp ph\u01b0\u01a1ng tr\u00ecnh bi\u1ec3u th\u1ecb l\u01b0\u1ee3ng c\u1ecf m\u1ed7i con \u0103n trong 1 tu\u1ea7n l\u00e0 nh\u01b0 nhau<br\/>B\u01b0\u1edbc 2: Gi\u1ea3i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh v\u1eeba l\u1eadp<br\/>B\u01b0\u1edbc 3: Ki\u1ec3m tra v\u00e0 k\u1ebft lu\u1eadn b\u00e0i to\u00e1n.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>B\u00e0i to\u00e1n kh\u00f3 \u1edf ch\u1ed7 trong su\u1ed1t th\u1eddi gian b\u00f2 \u0103n c\u1ecf, c\u1ecf v\u1eabn m\u1ecdc \u0111\u1ec1u tr\u00ean c\u00e1nh \u0111\u1ed3ng. <br\/>G\u1ecdi kh\u1ed1i l\u01b0\u1ee3ng c\u1ecf c\u00f3 s\u1eb5n tr\u00ean c\u00e1nh \u0111\u1ed3ng khi b\u00f2 \u0103n c\u1ecf l\u00e0 $1$ (\u0111\u01a1n v\u1ecb kh\u1ed1i l\u01b0\u1ee3ng quy \u01b0\u1edbc), kh\u1ed1i l\u01b0\u1ee3ng c\u1ecf m\u1ecdc th\u00eam tr\u00ean c\u00e1nh \u0111\u1ed3ng trong m\u1ed9t tu\u1ea7n l\u00e0 $y$ (v\u1edbi \u0111\u01a1n v\u1ecb kh\u1ed1i l\u01b0\u1ee3ng n\u00f3i tr\u00ean), $y>0$.<br\/> G\u1ecdi s\u1ed1 b\u00f2 ph\u1ea3i t\u00ecm l\u00e0 $x$ con, $x$ nguy\u00ean d\u01b0\u01a1ng.<br\/> $9$ con b\u00f2 \u0103n trong $2$ tu\u1ea7n h\u1ebft $1+2y$ n\u00ean m\u1ed7i con b\u00f2 trong $1$ tu\u1ea7n \u0103n h\u1ebft $\\dfrac{1+2y}{18}$<br\/> $6$ con b\u00f2 \u0103n trong $4$ tu\u1ea7n h\u1ebft $1+4y$ n\u00ean m\u1ed7i con b\u00f2 trong $1$ tu\u1ea7n \u0103n h\u1ebft $\\dfrac{1+4y}{24}$<br\/>$x$ con b\u00f2 \u0103n trong $6$ tu\u1ea7n h\u1ebft $1+6y$ n\u00ean m\u1ed7i con b\u00f2 trong $1$ tu\u1ea7n h\u1ebft $\\dfrac{1+6y}{6x}$<br\/> Ta c\u00f3 $\\dfrac{1+2y}{18}=\\dfrac{1+4y}{24}=\\dfrac{1+6y}{6x}$ .<br\/> Suy ra h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{aligned} & \\dfrac{1+2y}{18}=\\dfrac{1+4y}{24} \\\\ & \\dfrac{1+2y}{18}=\\dfrac{1+6y}{6x} \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & y=\\dfrac{1}{4} \\\\ & \\dfrac{1+2y}{18}=\\dfrac{1+6y}{6x} \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & y=\\dfrac{1}{4} \\\\ & x=5 \\\\ \\end{aligned} \\right.$<br\/>Suy ra c\u00f3 $5$ con b\u00f2 \u0103n h\u1ebft c\u1ecf c\u1ee7a c\u00e1nh \u0111\u1ed3ng trong $6$ tu\u1ea7n.<br\/><span class='basic_left'><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":4}]}],"id_ques":348},{"time":24,"part":[{"time":3,"title":"\u0110\u1ec3 s\u1eeda m\u1ed9t ng\u00f4i nh\u00e0 c\u1ea7n m\u1ed9t s\u1ed1 th\u1ee3 l\u00e0m vi\u1ec7c trong m\u1ed9t th\u1eddi gian quy \u0111\u1ecbnh. N\u1ebfu gi\u1ea3m ba ng\u01b0\u1eddi th\u00ec th\u1eddi gian k\u00e9o d\u00e0i th\u00eam $6$ ng\u00e0y, n\u1ebfu t\u0103ng th\u00eam hai ng\u01b0\u1eddi th\u00ec xong s\u1edbm h\u01a1n hai ng\u00e0y. H\u1ecfi theo quy \u0111\u1ecbnh th\u00ec c\u1ea7n bao nhi\u00eau th\u1ee3 v\u00e0 l\u00e0m trong bao nhi\u00eau ng\u00e0y \u0111\u1ec3 ho\u00e0n th\u00e0nh c\u00f4ng vi\u1ec7c tr\u00ean? Bi\u1ebft r\u1eb1ng kh\u1ea3 n\u0103ng lao \u0111\u1ed9ng c\u1ee7a m\u1ed7i th\u1ee3 \u0111\u1ec1u nh\u01b0 nhau?","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c l\u1eddi gi\u1ea3i \u0111\u00fang","temp":"sequence","correct":[[[3],[7],[1],[6],[5],[2],[4]]],"list":[{"point":10,"image":"img\/1.png","left":["N\u1ebfu gi\u1ea3m \u0111i $3$ ng\u01b0\u1eddi th\u00ec th\u1eddi gian k\u00e9o d\u00e0i th\u00eam $6$ ng\u00e0y. Nh\u01b0 v\u1eady $x-3$ ng\u01b0\u1eddi l\u00e0m trong $y+6$ ng\u00e0y th\u00ec \u0111\u01b0\u1ee3c $(x-3)(y+6)\\dfrac{1}{xy} =1$ (to\u00e0n b\u1ed9 c\u00f4ng vi\u1ec7c) (1)","V\u1eady theo quy \u0111\u1ecbnh th\u00ec c\u1ea7n $8$ th\u1ee3 v\u00e0 l\u00e0m trong $10$ ng\u00e0y \u0111\u1ec3 ho\u00e0n th\u00e0nh c\u00f4ng vi\u1ec7c tr\u00ean.","G\u1ecdi s\u1ed1 th\u1ee3 c\u1ea7n thi\u1ebft theo quy \u0111\u1ecbnh l\u00e0 $x$ (ng\u01b0\u1eddi), $x\\in {{\\mathbb{N}}^{*}}$ , th\u1eddi gian c\u1ea7n thi\u1ebft l\u00e0 $y$ (ng\u00e0y), $y>0.$ ","Gi\u1ea3i h\u1ec7 ta \u0111\u01b0\u1ee3c $\\left( x;y \\right)=\\left( 8;10 \\right)$ (th\u1ecfa m\u00e3n)","T\u1eeb (1) v\u00e0 (2), ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: $\\left\\{ \\begin{align} & \\left( x-3 \\right)\\left( y+6 \\right)~\\dfrac{1}{xy}=1 \\\\ & \\left( x+2 \\right)\\left( y-2 \\right)\\dfrac{1}{xy}=1 \\\\ \\end{align} \\right.$","Coi to\u00e0n b\u1ed9 c\u00f4ng vi\u1ec7c nh\u01b0 m\u1ed9t \u0111\u01a1n v\u1ecb c\u00f4ng vi\u1ec7c th\u00ec m\u1ed9t ng\u01b0\u1eddi th\u1ee3 trong ng\u00e0y l\u00e0m \u0111\u01b0\u1ee3c $\\dfrac{1}{xy}$ (c\u00f4ng vi\u1ec7c)","T\u01b0\u01a1ng t\u1ef1 n\u1ebfu t\u0103ng th\u00eam hai ng\u01b0\u1eddi th\u00ec ch\u1ec9 c\u1ea7n $y-2$ ng\u00e0y. Nh\u01b0 v\u1eady $x+2$ ng\u01b0\u1eddi l\u00e0m trong $y-2$ ng\u00e0y \u0111\u01b0\u1ee3c $\\left( x+2 \\right)\\left( y-2 \\right)\\dfrac{1}{xy}=1$ (to\u00e0n b\u1ed9 c\u00f4ng vi\u1ec7c) (2)"],"top":105,"hint":"Coi to\u00e0n b\u1ed9 c\u00f4ng vi\u1ec7c nh\u01b0 m\u1ed9t \u0111\u01a1n v\u1ecb c\u00f4ng vi\u1ec7c <br\/>Ch\u00fa \u00fd: S\u1ed1 ph\u1ea7n c\u00f4ng vi\u1ec7c trong m\u1ed9t ng\u00e0y v\u00e0 th\u1eddi gian c\u1ea7n thi\u1ebft \u0111\u1ec3 ho\u00e0n th\u00e0nh c\u00f4ng vi\u1ec7c l\u00e0 hai \u0111\u1ea1i l\u01b0\u1ee3ng t\u1ec9 l\u1ec7 ngh\u1ecbch.","explain":"<span class='basic_left'>G\u1ecdi s\u1ed1 th\u1ee3 c\u1ea7n thi\u1ebft theo quy \u0111\u1ecbnh l\u00e0 $x$ (ng\u01b0\u1eddi), $x\\in {{\\mathbb{N}}^{*}}$ , th\u1eddi gian c\u1ea7n thi\u1ebft l\u00e0 $y$ (ng\u00e0y), $y>0.$<br\/>Coi to\u00e0n b\u1ed9 c\u00f4ng vi\u1ec7c nh\u01b0 m\u1ed9t \u0111\u01a1n v\u1ecb c\u00f4ng vi\u1ec7c th\u00ec m\u1ed9t ng\u01b0\u1eddi th\u1ee3 trong ng\u00e0y l\u00e0m \u0111\u01b0\u1ee3c $\\dfrac{1}{xy}$ (c\u00f4ng vi\u1ec7c)<br\/>N\u1ebfu gi\u1ea3m \u0111i $3$ ng\u01b0\u1eddi th\u00ec th\u1eddi gian k\u00e9o d\u00e0i th\u00eam $6$ ng\u00e0y. Nh\u01b0 v\u1eady $x-3$ ng\u01b0\u1eddi l\u00e0m trong $y+6$ ng\u00e0y th\u00ec \u0111\u01b0\u1ee3c $(x-3)(y+6)\\dfrac{1}{xy} =1$ (to\u00e0n b\u1ed9 c\u00f4ng vi\u1ec7c)(1)<br\/>T\u01b0\u01a1ng t\u1ef1 n\u1ebfu t\u0103ng th\u00eam hai ng\u01b0\u1eddi th\u00ec ch\u1ec9 c\u1ea7n $y-2$ ng\u00e0y. Nh\u01b0 v\u1eady $x+2$ ng\u01b0\u1eddi l\u00e0m trong $y-2$ ng\u00e0y \u0111\u01b0\u1ee3c $\\left( x+2 \\right)\\left( y-2 \\right)\\dfrac{1}{xy}=1$ (to\u00e0n b\u1ed9 c\u00f4ng vi\u1ec7c)(2)<br\/>Ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{align} & \\left( x-3 \\right)\\left( y+6 \\right)~\\dfrac{1}{xy}=1 \\\\ & \\left( x+2 \\right)\\left( y-2 \\right)\\dfrac{1}{xy}=1 \\\\ \\end{align} \\right.$<br\/>$\\Leftrightarrow \\left\\{ \\begin{align} & \\left( x-3 \\right)\\left( y+6 \\right)=xy \\\\ & \\left( x+2 \\right)\\left( y-2 \\right)=xy \\\\ \\end{align} \\right.$<br\/> Gi\u1ea3i h\u1ec7 ta \u0111\u01b0\u1ee3c $\\left( x;y \\right)=\\left( 8;10 \\right)$(th\u1ecfa m\u00e3n)<br\/> V\u1eady theo quy \u0111\u1ecbnh th\u00ec c\u1ea7n $8$ th\u1ee3 v\u00e0 l\u00e0m trong $10$ ng\u00e0y \u0111\u1ec3 ho\u00e0n th\u00e0nh c\u00f4ng vi\u1ec7c tr\u00ean.<\/span>"}]}],"id_ques":349},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["30"],["20"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>Trong m\u1ed9t trang s\u00e1ch, n\u1ebfu b\u1edbt \u0111i $5$ d\u00f2ng v\u00e0 m\u1ed7i d\u00f2ng b\u1edbt \u0111i $2$ ch\u1eef th\u00ec c\u1ea3 trang b\u1edbt \u0111i $150$ ch\u1eef, n\u1ebfu t\u0103ng th\u00eam $2$ d\u00f2ng v\u00e0 m\u1ed7i d\u00f2ng th\u00eam $6$ ch\u1eef th\u00ec c\u1ea3 trang s\u1ebd t\u0103ng th\u00eam $232$ ch\u1eef. T\u00ednh s\u1ed1 d\u00f2ng trong trang v\u00e0 s\u1ed1 ch\u1eef c\u00f3 trong m\u1ed7i d\u00f2ng<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> S\u1ed1 d\u00f2ng trong trang l\u00e0 _input_. S\u1ed1 ch\u1eef trong 1 d\u00f2ng l\u00e0 _input_ <\/span>","explain":"<span class='basic_left'>G\u1ecdi $x$ l\u00e0 s\u1ed1 d\u00f2ng trong $1$ trang v\u00e0 $y$ l\u00e0 s\u1ed1 ch\u1eef tr\u00ean $1$ d\u00f2ng $(x,y\\in \\mathbb{N^*})$ <br\/> Khi \u0111\u00f3, c\u1ea3 trang c\u00f3 $xy$ ch\u1eef <br\/> V\u00ec b\u1edbt \u0111i $2$ d\u00f2ng v\u00e0 m\u1ed7i d\u00f2ng b\u1edbt \u0111i $2$ ch\u1eef th\u00ec c\u1ea3 trang b\u1edbt \u0111i $150$ ch\u1eef n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: <\/span> $\\left( x-5 \\right)\\left( y-2 \\right)=xy-150$ (1) <br\/><span class='basic_left'>V\u00ec t\u0103ng th\u00eam $2$ d\u00f2ng v\u00e0 m\u1ed7i d\u00f2ng th\u00eam $6$ ch\u1eef th\u00ec c\u1ea3 trang s\u1ebd t\u0103ng th\u00eam $232$ ch\u1eef n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: <\/span> $\\left( x+2 \\right)\\left( y+6 \\right)=xy+232$ (2)<br\/><span class='basic_left'>T\u1eeb (1) v\u00e0 (2), ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh$\\left\\{ \\begin{align} & \\left( x-5 \\right)\\left( y-2 \\right)=xy-150 \\\\ & \\left( x+2 \\right)\\left( y+6 \\right)=xy+232 \\\\ \\end{align} \\right.$ <br\/>$\\Leftrightarrow \\left\\{ \\begin{aligned} & 2x+5y=160 \\\\ & 6x+2y=220 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & 2x+5y=160 \\\\ & 3x+y=110 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & 2x+5y=160 \\\\ & 15x+5y=550 \\\\ \\end{aligned} \\right. \\Leftrightarrow \\left\\{ \\begin{aligned} & 13x=390 \\\\ & 3x+y=110 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x=30 \\\\ & y=20 \\\\ \\end{aligned} \\right.$ <br\/>V\u1eady m\u1ed7i trang c\u00f3 $30$ d\u00f2ng v\u00e0 m\u1ed7i d\u00f2ng $20$ ch\u1eef.<br\/><span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 l\u00e0 $30$ v\u00e0 $20$<\/span><\/span>"}]}],"id_ques":350}],"lesson":{"save":0,"level":3}}