{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["20"]]],"list":[{"point":5,"width":60,"type_input":"","ques":"<span class='basic_left'>Hai \u0111\u1ed9i c\u00f4ng nh\u00e2n c\u00f9ng l\u00e0m m\u1ed9t c\u00f4ng vi\u1ec7c th\u00ec $12$ ng\u00e0y xong vi\u1ec7c. N\u1ebfu \u0111\u1ed9i th\u1ee9 nh\u1ea5t l\u00e0m m\u1ed9t m\u00ecnh h\u1ebft n\u1eeda c\u00f4ng vi\u1ec7c r\u1ed3i \u0111\u1ed9i th\u1ee9 hai ti\u1ebfp t\u1ee5c m\u1ed9t m\u00ecnh l\u00e0m n\u1ed1t ph\u1ea7n c\u00f4ng vi\u1ec7c c\u00f2n l\u1ea1i th\u00ec h\u1ebft t\u1ea5t c\u1ea3 $25$ ng\u00e0y. H\u1ecfi th\u1eddi gian \u0111\u1ed9i th\u1ee9 nh\u1ea5t l\u00e0m m\u1ed9t m\u00ecnh xong c\u00f4ng vi\u1ec7c bao nhi\u00eau ng\u00e0y bi\u1ebft \u0111\u1ed9i th\u1ee9 nh\u1ea5t l\u00e0m m\u1ed9t m\u00ecnh nhanh h\u01a1n \u0111\u1ed9i th\u1ee9 hai l\u00e0m m\u1ed9t m\u00ecnh<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> _input_ (ng\u00e0y)","hint":"G\u1ecdi th\u1eddi gian \u0111\u1ed9i th\u1ee9 nh\u1ea5t l\u00e0m m\u1ed9t m\u00ecnh xong n\u1eeda c\u00f4ng vi\u1ec7c l\u00e0 $x$ ng\u00e0y, khi \u0111\u00f3 th\u1eddi gian \u0111\u1ed9i th\u1ee9 hai l\u00e0m m\u1ed9t m\u00ecnh xong n\u1eeda c\u00f4ng vi\u1ec7c l\u00e0 $25-x$ (ng\u00e0y)","explain":"<span class='basic_left'>G\u1ecdi th\u1eddi gian \u0111\u1ed9i th\u1ee9 nh\u1ea5t l\u00e0m m\u1ed9t m\u00ecnh xong n\u1eeda c\u00f4ng vi\u1ec7c l\u00e0 $x$ ng\u00e0y.<br\/>Th\u1eddi gian \u0111\u1ed9i th\u1ee9 hai l\u00e0m m\u1ed9t m\u00ecnh xong n\u1eeda c\u00f4ng vi\u1ec7c l\u00e0 $25-x$ (ng\u00e0y).<br\/>\u0110i\u1ec1u ki\u1ec7n: $\\left\\{ \\begin{align} & 2x > 12 \\\\ & x < 25 \\\\ & x < 25-x \\\\ \\end{align} \\right.\\Leftrightarrow 6 < x < \\dfrac{25}{2}$ <br\/>Suy ra th\u1eddi gian \u0111\u1ed9i th\u1ee9 nh\u1ea5t v\u00e0 \u0111\u1ed9i th\u1ee9 hai l\u00e0m m\u1ed9t m\u00ecnh xong c\u1ea3 c\u00f4ng vi\u1ec7c l\u00e0 $2x$ (ng\u00e0y) v\u00e0 $2(25-x)$ (ng\u00e0y)<br\/>Trong m\u1ed9t ng\u00e0y, \u0111\u1ed9i th\u1ee9 nh\u1ea5t l\u00e0m \u0111\u01b0\u1ee3c $\\dfrac{1}{2x}$ (c\u00f4ng vi\u1ec7c)<br\/>Trong m\u1ed9t ng\u00e0y, \u0111\u1ed9i th\u1ee9 hai l\u00e0m \u0111\u01b0\u1ee3c $\\dfrac{1}{2\\left( 25-x \\right)}$ (c\u00f4ng vi\u1ec7c)<br\/>Trong m\u1ed9t ng\u00e0y, c\u1ea3 hai \u0111\u1ed9i l\u00e0m \u0111\u01b0\u1ee3c $\\dfrac{1}{12}$ (c\u00f4ng vi\u1ec7c)<br\/>Ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\dfrac{1}{2x}+\\dfrac{1}{2\\left( 25-x \\right)}=\\dfrac{1}{12}.$<br\/>$\\begin{align} & \\Leftrightarrow \\dfrac{1}{x}+\\dfrac{1}{25-x}=\\dfrac{1}{6} \\\\ & \\Rightarrow 6\\left( 25-x \\right)+6x=x\\left( 25-x \\right) \\\\ & \\Leftrightarrow {{x}^{2}}-25x+150=0 \\\\ \\end{align}$<br\/> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c $x_1=10;x_2=15$<br\/>V\u00ec $ 6 < x < \\dfrac{25}{2} $ n\u00ean $x=x_1=10$. Suy ra $2x=20$<br\/>V\u1eady th\u1eddi gian \u0111\u1ed9i th\u1ee9 nh\u1ea5t l\u00e0m m\u1ed9t m\u00ecnh xong c\u1ea3 c\u00f4ng vi\u1ec7c l\u00e0 $20$ ng\u00e0y.<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $20.$<\/span><\/span>"}]}],"id_ques":1011},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng","title_trans":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n d\u01b0\u1edbi d\u1ea1ng s\u1ed1 th\u1eadp ph\u00e2n","temp":"fill_the_blank","correct":[[["0,5"]]],"list":[{"point":5,"width":60,"type_input":"","ques":"<span class='basic_left'>Sau $2$ n\u0103m, s\u1ed1 d\u00e2n c\u1ee7a m\u1ed9t th\u00e0nh ph\u1ed1 t\u0103ng t\u1eeb $2\\,000\\,000$ ng\u01b0\u1eddi l\u00ean $2\\, 020\\, 050$ ng\u01b0\u1eddi. H\u1ecfi trung b\u00ecnh m\u1ed7i n\u0103m d\u00e2n s\u1ed1 c\u1ee7a th\u00e0nh ph\u1ed1 t\u0103ng bao nhi\u00eau $\\%.$?<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> _input_$\\%$","hint":"G\u1ecdi t\u1ec9 l\u1ec7 t\u0103ng d\u00e2n s\u1ed1 trung b\u00ecnh m\u1ed7i n\u0103m l\u00e0 $x\\%$. D\u00e2n s\u1ed1 c\u1ee7a th\u00e0nh ph\u1ed1 sau $1$ n\u0103m l\u00e0 $N+N.x\\%$ (ng\u01b0\u1eddi), trong \u0111\u00f3 $N$ l\u00e0 d\u00e2n s\u1ed1 ban \u0111\u1ea7u","explain":"<span class='basic_left'>G\u1ecdi t\u1ec9 l\u1ec7 t\u0103ng d\u00e2n s\u1ed1 trung b\u00ecnh m\u1ed7i n\u0103m l\u00e0 $x\\%,$ $ x>0$<br\/>Sau m\u1ed9t n\u0103m, d\u00e2n s\u1ed1 c\u1ee7a th\u00e0nh ph\u1ed1 l\u00e0:<br\/>$2\\,000\\,000+2\\,000\\,000.\\dfrac{x}{100}=2\\,000\\,000+20\\,000x$(ng\u01b0\u1eddi)<br\/>Sau hai n\u0103m, d\u00e2n s\u1ed1 c\u1ee7a th\u00e0nh ph\u1ed1 l\u00e0:<br\/> $2\\,000\\,000+20\\,000x+\\left( 2\\,000\\,000+20\\,000x \\right)\\dfrac{x}{100}=2\\,000\\,000+40\\,000x+200{{x}^{2}}$(ng\u01b0\u1eddi)<br\/>Do sau hai n\u0103m d\u00e2n s\u1ed1 l\u00e0 $2\\, 020\\, 050$ ng\u01b0\u1eddi n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$2\\,000\\,000+40\\,000x+200{{x}^{2}}=2\\,020\\,050$<br\/>Hay $4{{x}^{2}}+800x-401=0$ <br\/> $\\Delta '=160\\,000+1604=161\\,604\\Rightarrow \\sqrt{\\Delta '}=402$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0:<br\/>${{x}_{1}}=\\dfrac{-400+402}{4}=0,5;{{x}_{2}}=\\dfrac{-400-402}{4}<0$ <br\/>V\u00ec $x>0$ n\u00ean ${{x}_{2}}$ lo\u1ea1i, ${{x}_{1}}$ th\u1ecfa m\u00e3n<br\/>Do \u0111\u00f3 trung b\u00ecnh m\u1ed7i n\u0103m d\u00e2n s\u1ed1 c\u1ee7a th\u00e0nh ph\u1ed1 t\u0103ng $0,5 \\%.$<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0,5.$<\/span><\/span>"}]}],"id_ques":1012},{"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":[[["20"],["24"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"Trong c\u00f9ng m\u1ed9t th\u1eddi gian nh\u01b0 nhau, \u0111\u1ed9i I ph\u1ea3i \u0111\u00e0o $810$ $m^3 $ \u0111\u1ea5t, \u0111\u1ed9i II ph\u1ea3i \u0111\u00e0o $900$ $m^3 $ \u0111\u1ea5t. K\u1ebft qu\u1ea3 \u0111\u1ed9i I \u0111\u00e3 ho\u00e0n th\u00e0nh tr\u01b0\u1edbc th\u1eddi h\u1ea1n $3$ ng\u00e0y, \u0111\u1ed9i II ho\u00e0n th\u00e0nh tr\u01b0\u1edbc th\u1eddi h\u1ea1n $6 $ ng\u00e0y. T\u00ednh s\u1ed1 \u0111\u1ea5t m\u1ed7i \u0111\u1ed9i \u0111\u00e3 \u0111\u00e0o \u0111\u01b0\u1ee3c trong m\u1ed9t ng\u00e0y bi\u1ebft r\u1eb1ng m\u1ed7i ng\u00e0y \u0111\u1ed9i II \u0111\u00e3 \u0111\u00e0o nhi\u00eau h\u01a1n \u0111\u1ed9i I l\u00e0 $4$ $m^3.$<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> \u0110\u1ed9i I: _input_$(m^3\/$ng\u00e0y); \u0111\u1ed9i II: _input_$(m^3\/$ng\u00e0y)","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/>L\u1eadp b\u1ea3ng: <br\/> <table><tr><th>C\u00e1c \u0111\u1ea1i l\u01b0\u1ee3ng<br><\/th><th>S\u1ed1 \u0111\u1ea5t c\u1ea7n \u0111\u00e0o <br\/>$(m^3)$<br><\/th><th>S\u1ed1 \u0111\u1ea5t \u0111\u00e0o m\u1ed7i ng\u00e0y<br\/>($m^3\/$ng\u00e0y)<br><\/th><th>Th\u1eddi gian <br\/>(ng\u00e0y)<br><\/th><\/tr><tr><th>\u0110\u1ed9i I<br><\/th><td>$810$<\/td><td>$x$<\/td><td>$\\dfrac{810}{x}$<\/td><\/tr><tr><th>\u0110\u1ed9i II<br><\/th><td>$900$<\/td><td>$x+4$<\/td><td>$\\dfrac{900}{x+4}$<\/td><\/tr><\/table><br\/>V\u00ec hai \u0111\u1ed9i ph\u1ea3i \u0111\u00e0o trong c\u00f9ng m\u1ed9t th\u1eddi gian nh\u01b0 nhau, nh\u01b0ng \u0111\u1ed9i I \u0111\u00e3 ho\u00e0n th\u00e0nh tr\u01b0\u1edbc th\u1eddi h\u1ea1n $3$ ng\u00e0y, \u0111\u1ed9i II ho\u00e0n th\u00e0nh tr\u01b0\u1edbc th\u1eddi h\u1ea1n $6$ ng\u00e0y n\u00ean th\u1eddi gian \u0111\u1ed9i I ho\u00e0n th\u00e0nh nhi\u1ec1u h\u01a1n th\u1eddi gian \u0111\u1ed9i II ho\u00e0n th\u00e0nh l\u00e0 $3$ ng\u00e0y. C\u0103n c\u1ee9 v\u00e0o \u0111\u00f3 \u0111\u1ec3 l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>G\u1ecdi s\u1ed1 \u0111\u1ea5t \u0111\u1ed9i I \u0111\u00e3 \u0111\u00e0o m\u1ed9t ng\u00e0y l\u00e0 $x$ $(m^3),$ $x>0.$<br\/>V\u00ec m\u1ed7i ng\u00e0y \u0111\u1ed9i II \u0111\u00e3 \u0111\u00e0o nhi\u00eau h\u01a1n \u0111\u1ed9i I l\u00e0 $4$ $m^3$ n\u00ean s\u1ed1 \u0111\u1ea5t \u0111\u1ed9i II \u0111\u00e0o m\u1ed9t ng\u00e0y l\u00e0 $x+4 $ $(m^3).$<br\/>Th\u1eddi gian \u0111\u1ed9i I \u0111\u00e0o $810$ $m^3$ \u0111\u1ea5t l\u00e0 $\\dfrac{810}{x}$ ng\u00e0y; th\u1eddi gian \u0111\u1ed9i II \u0111\u00e0o $900$ $m^3 $ \u0111\u1ea5t l\u00e0 $\\dfrac{900}{x+4}$ $(m^3).$<br\/>V\u00ec hai \u0111\u1ed9i ph\u1ea3i \u0111\u00e0o trong c\u00f9ng m\u1ed9t th\u1eddi gian nh\u01b0 nhau, nh\u01b0ng \u0111\u1ed9i I \u0111\u00e3 ho\u00e0n th\u00e0nh tr\u01b0\u1edbc th\u1eddi h\u1ea1n $3$ ng\u00e0y, \u0111\u1ed9i II ho\u00e0n th\u00e0nh tr\u01b0\u1edbc th\u1eddi h\u1ea1n $6$ ng\u00e0y n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: <br\/>$\\begin{align} & \\,\\,\\,\\,\\,\\,\\dfrac{810}{x}-\\dfrac{900}{x+4}=6-3 \\\\ & \\Leftrightarrow \\,\\dfrac{810}{x}-\\dfrac{900}{x+4}=3 \\\\ & \\Rightarrow 810\\left( x+4 \\right)-900x=3x\\left( x+4 \\right) \\\\ & \\Leftrightarrow {{x}^{2}}+34x-1080=0 \\\\ \\end{align}$ <br\/>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c: $x_1=20$ (th\u1ecfa m\u00e3n); $x_2=-54$ (lo\u1ea1i)<br\/>V\u1eady \u0111\u1ed9i I \u0111\u00e0o $20$ $m^3$ m\u1ed9t ng\u00e0y, \u0111\u1ed9i II \u0111\u00e0o $24$ $m^3$ m\u1ed9t ng\u00e0y.<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 $20$ v\u00e0 $24.$ <\/span><\/span>"}]}],"id_ques":1013},{"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":[[["24"],["18"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>Hai ng\u01b0\u1eddi \u0111i xe \u0111\u1ea1p c\u00f9ng kh\u1edfi h\u00e0nh m\u1ed9t l\u00fac \u1edf c\u00f9ng m\u1ed9t ch\u1ed7. Ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t \u0111i v\u1ec1 ph\u00eda B\u1eafc, ng\u01b0\u1eddi th\u1ee9 hai \u0111i v\u1ec1 ph\u00eda \u0110\u00f4ng. Sau $2$ gi\u1edd, h\u1ecd c\u00e1ch nhau $60$ $km$ theo \u0111\u01b0\u1eddng chim bay. Bi\u1ebft v\u1eadn t\u1ed1c c\u1ee7a ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t l\u1edbn h\u01a1n v\u1eadn t\u1ed1c c\u1ee7a ng\u01b0\u1eddi th\u1ee9 hai l\u00e0 $6$ $km\/h.$ T\u00ednh v\u1eadn t\u1ed1c m\u1ed7i ng\u01b0\u1eddi.<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> Ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t: _input_$km\/h$; ng\u01b0\u1eddi th\u1ee9 hai: _input_ $km\/h$<\/span>","hint":"Ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t \u0111i v\u1ec1 ph\u00eda B\u1eafc, ng\u01b0\u1eddi th\u1ee9 hai \u0111i v\u1ec1 ph\u00eda \u0110\u00f4ng n\u00ean h\u01b0\u1edbng chuy\u1ebfn \u0111\u1ed9ng c\u1ee7a hai ng\u01b0\u1eddi vu\u00f4ng g\u00f3c v\u1edbi nhau","explain":"<span class='basic_left'><center><img src= 'https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai23/lv3/img\/K23.2a.png' \/><\/center>G\u1ecdi v\u1eadn t\u1ed1c c\u1ee7a ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t l\u00e0 $x$ $(km\/h).$<br\/>V\u00ec v\u1eadn t\u1ed1c c\u1ee7a ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t l\u1edbn h\u01a1n v\u1eadn t\u1ed1c c\u1ee7a ng\u01b0\u1eddi th\u1ee9 hai l\u00e0 $6$ $km\/h$ n\u00ean v\u1eadn t\u1ed1c c\u1ee7a ng\u01b0\u1eddi th\u1ee9 hai l\u00e0 $x-6$ $(km\/h),$ $x>6$<br\/>Sau $2$ gi\u1edd, ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t \u0111i \u0111\u01b0\u1ee3c $2x$ $(km)$ v\u00e0 ng\u01b0\u1eddi th\u1ee9 hai \u0111i \u0111\u01b0\u1ee3c $2(x-6)$ $km.$<br\/>V\u00ec ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t \u0111i v\u1ec1 ph\u00eda B\u1eafc, ng\u01b0\u1eddi th\u1ee9 hai \u0111i v\u1ec1 ph\u00eda \u0110\u00f4ng n\u00ean h\u01b0\u1edbng chuy\u1ebfn \u0111\u1ed9ng c\u1ee7a hai ng\u01b0\u1eddi vu\u00f4ng g\u00f3c v\u1edbi nhau nh\u01b0 h\u00ecnh v\u1ebd.<br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pytago cho tam gi\u00e1c vu\u00f4ng $ABC$, ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: <br\/>$(2x)^2+(2x-12)^2=60^2$<br\/>$\\begin{align} & \\Leftrightarrow 4{{x}^{2}}+4{{x}^{2}}-48x+144=3600 \\\\ & \\Leftrightarrow 8{{x}^{2}}-48x-3456=0 \\\\ & \\Leftrightarrow {{x}^{2}}-6x-432=0 \\\\ \\end{align}$ <br\/>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c ${{x}_{1}}=24;{{x}_{2}}=-18$ (lo\u1ea1i)<br\/>V\u1eady v\u1eadn t\u1ed1c c\u1ee7a ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t l\u00e0 $24$ $(km\/h),$ v\u1eadn t\u1ed1c c\u1ee7a ng\u01b0\u1eddi th\u1ee9 hai l\u00e0 $18$ $(km\/h)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 $24$ v\u00e0 $18.$ <\/span><\/span>"}]}],"id_ques":1014},{"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":[[["50"],["40"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>M\u1ed9t tam gi\u00e1c c\u00f3 chi\u1ec1u cao b\u1eb1ng $\\dfrac{3}{4}$ c\u1ea1nh \u0111\u00e1y t\u01b0\u01a1ng \u1ee9ng. N\u1ebfu chi\u1ec1u cao t\u0103ng th\u00eam $3 $ $dm,$ c\u1ea1nh \u0111\u00e1y gi\u1ea3m $2$ $dm$ th\u00ec di\u1ec7n t\u00edch t\u0103ng th\u00eam $8$$\\%.$ T\u00ednh chi\u1ec1u cao v\u00e0 c\u1ea1nh \u0111\u00e1y c\u1ee7a tam gi\u00e1c bi\u1ebft c\u1ea1nh \u0111\u00e1y c\u00f3 \u0111\u1ed9 d\u00e0i l\u1edbn h\u01a1n $10$ $dm.$<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> C\u1ea1nh \u0111\u00e1y: _input_ $(dm)$; chi\u1ec1u cao: _input_$(dm)$<\/span>","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/>L\u1eadp b\u1ea3ng: <br\/> <table><tr><th>C\u00e1c \u0111\u1ea1i l\u01b0\u1ee3ng<br><\/th><th>C\u1ea1nh \u0111\u00e1y $(dm)$<br><\/th><th>Chi\u1ec1u cao $(dm)$<br><\/th><th>Di\u1ec7n t\u00edch $(dm^2)$<br><\/th><\/tr><tr><th>Ban \u0111\u1ea7u<br><\/th><td>$x$<\/td><td>$\\dfrac{3}{4}x$<\/td><td>$\\dfrac{1}{2}.\\dfrac{3}{4}x.x$<\/td><\/tr><tr><th>Sau khi thay \u0111\u1ed5i<br><\/th><td>$x-2$<\/td><td>$\\dfrac{3}{4}x+3$<\/td><td> $\\dfrac{1}{2}\\left( \\dfrac{3}{4}x+3 \\right)\\left( x-2 \\right)$<\/td><\/tr><\/table><br\/>C\u0103n c\u1ee9 v\u00e0o gi\u1ea3 thi\u1ebft: Di\u1ec7n t\u00edch tam gi\u00e1c m\u1edbi t\u0103ng th\u00eam $8\\%$ so v\u1edbi di\u1ec7n t\u00edch c\u0169, \u0111\u1ec3 l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>G\u1ecdi c\u1ea1nh \u0111\u00e1y c\u1ee7a tam gi\u00e1c l\u00e0 $x$ $(dm),$ $x>10$<br\/>Chi\u1ec1u cao c\u1ee7a tam gi\u00e1c l\u00e0 $\\dfrac{3}{4}x$ $(dm)$<br\/>Di\u1ec7n t\u00edch ban \u0111\u1ea7u c\u1ee7a tam gi\u00e1c l\u00e0 $\\dfrac{1}{2}.\\dfrac{3}{4}x.x=\\dfrac{3}{8}{{x}^{2}}$ $(dm^2)$<br\/>N\u1ebfu chi\u1ec1u cao t\u0103ng th\u00eam $3$ $dm$ v\u00e0 c\u1ea1nh \u0111\u00e1y gi\u1ea3m $2$ $dm$ th\u00ec di\u1ec7n t\u00edch m\u1edbi c\u1ee7a tam gi\u00e1c l\u00e0 $\\dfrac{1}{2}\\left( \\dfrac{3}{4}x+3 \\right)\\left( x-2 \\right)$ $(dm^2)$<br\/>Di\u1ec7n t\u00edch tam gi\u00e1c m\u1edbi t\u0103ng th\u00eam $8\\%$ so v\u1edbi di\u1ec7n t\u00edch c\u0169 n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\begin{aligned} & \\,\\,\\,\\,\\,\\,\\,\\dfrac{1}{2}\\left( \\dfrac{3}{4}x+3 \\right)\\left( x-2 \\right)=\\left( 100+8 \\right)\\%.\\dfrac{3}{8}{{x}^{2}} \\\\ & \\Leftrightarrow \\dfrac{1}{2}\\left( \\dfrac{3}{4}{{x}^{2}}+3x-\\dfrac{3}{2}x-6 \\right)=\\dfrac{108}{100}.\\dfrac{3}{8}{{x}^{2}} \\\\ & \\Leftrightarrow \\dfrac{3}{8}{{x}^{2}}+\\dfrac{3}{4}x-3=\\dfrac{81}{200}{{x}^{2}} \\\\ & \\Leftrightarrow \\dfrac{3}{100}{{x}^{2}}-\\dfrac{3}{4}x+3=0 \\\\ & \\Leftrightarrow {{x}^{2}}-25x+100=0 \\\\ & \\Delta =225\\Rightarrow \\sqrt{\\Delta }=15 \\\\ & \\Rightarrow {{x}_{1}}=\\dfrac{25+15}{2}=20;{{x}_{2}}=\\dfrac{25-15}{2}=5 \\\\ \\end{aligned}$ <br\/>V\u00ec $ x > 10 $ n\u00ean $ x=x_1=20 $<br\/>V\u1eady c\u1ea1nh \u0111\u00e1y c\u1ee7a tam gi\u00e1c l\u00e0 $20$ $(dm),$ chi\u1ec1u cao c\u1ee7a tam gi\u00e1c l\u00e0 $\\dfrac{3}{4}x=\\dfrac{3}{4}.20=15 $ $(dm)$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 $20$ v\u00e0 $15.$ <\/span><\/span>"}]}],"id_ques":1015},{"time":24,"part":[{"title":"Ch\u1ecdn <u>nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["1","4"]],"list":[{"point":5,"img":"","ques":"<span class='basic_left'>M\u1ed9t h\u1ee3p kim g\u1ed3m \u0111\u1ed3ng v\u00e0 k\u1ebdm, trong \u0111\u00f3 c\u00f3 $5$ $kg$ k\u1ebdm. N\u1ebfu th\u00eam $15$$ kg$ k\u1ebdm v\u00e0o h\u1ee3p kim n\u00e0y th\u00ec ta \u0111\u01b0\u1ee3c h\u1ee3p kim m\u1edbi. K\u1ebft qu\u1ea3 trong h\u1ee3p kim m\u1edbi t\u1ec9 l\u1ec7 \u0111\u1ed3ng gi\u1ea3m $30$$\\%.$ Kh\u1ed1i l\u01b0\u1ee3ng ban \u0111\u1ea7u c\u1ee7a h\u1ee3p kim l\u00e0:<\/span>","hint":"","column":2,"number_true":2,"select":["A. $10$ $kg$","B. $15$ $kg$","C. $20$ $kg$ ","D. $25$ $kg$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/>L\u1eadp b\u1ea3ng: <br\/> <table><tr><th>C\u00e1c \u0111\u1ea1i l\u01b0\u1ee3ng<br><\/th><th>Kh\u1ed1i l\u01b0\u1ee3ng h\u1ee3p kim $(kg)$<br><\/th><th>Kh\u1ed1i l\u01b0\u1ee3ng \u0111\u1ed3ng $(kg)$<br><\/th><th>T\u1ec9 l\u1ec7 \u0111\u1ed3ng<br><\/th><\/tr><tr><th>Ban \u0111\u1ea7u<br><\/th><td>$x$<\/td><td>$x-5$<\/td><td>$\\dfrac{x-5}{x}$<\/td><\/tr><tr><th>L\u00fac sau<br><\/th><td>$x+15$<\/td><td>$x-5$<\/td><td> $\\dfrac{x-5}{x+15}$<\/td><\/tr><\/table><br\/>L\u1eadp ph\u01b0\u01a1ng tr\u00ecnh bi\u1ec3u th\u1ecb trong h\u1ee3p kim m\u1edbi t\u1ec9 l\u1ec7 \u0111\u1ed3ng gi\u1ea3m $30\\%.$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>G\u1ecdi kh\u1ed1i l\u01b0\u1ee3ng h\u1ee3p kim l\u00fac \u0111\u1ea7u l\u00e0 $x$ $(kg), $$x>0.$<br\/>Kh\u1ed1i l\u01b0\u1ee3ng \u0111\u1ed3ng c\u00f3 trong h\u1ee3p kim l\u00fac \u0111\u1ea7u l\u00e0 $x-5$ $(kg),$ $x>5$<br\/>T\u1ec9 l\u1ec7 \u0111\u1ed3ng c\u00f3 trong h\u1ee3p kim l\u00fac \u0111\u1ea7u l\u00e0 $\\dfrac{x-5}{x}$ <br\/>Kh\u1ed1i l\u01b0\u1ee3ng h\u1ee3p kim l\u00fac sau l\u00e0 $x+15$ $(kg)$<br\/>T\u1ec9 l\u1ec7 \u0111\u1ed3ng trong h\u1ee3p kim l\u00fac sau l\u00e0 $\\dfrac{x-5}{x+15}$<br\/>Do trong h\u1ee3p kim m\u1edbi t\u1ec9 l\u1ec7 \u0111\u1ed3ng gi\u1ea3m $30$ $\\%$ n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\begin{align} & \\,\\,\\,\\,\\,\\dfrac{x-5}{x}-\\dfrac{x-5}{x+15}=\\dfrac{3}{10} \\\\ & \\Rightarrow 10\\left( x-5 \\right)\\left( x+15 \\right)-10x\\left( x-5 \\right)=3x\\left( x+15 \\right) \\\\ & \\Leftrightarrow 10{{x}^{2}}-50x+150x-750-10{{x}^{2}}+50x=3{{x}^{2}}+45x \\\\ & \\Leftrightarrow {{x}^{2}}-35x+250=0 \\\\ & \\Delta =225\\Rightarrow \\sqrt{\\Delta }=15 \\\\ & \\Rightarrow {{x}_{1}}=\\dfrac{35+15}{2}=25;{{x}_{2}}=\\dfrac{35-15}{2}=10 \\\\ \\end{align}$<br\/> C\u1ea3 hai gi\u00e1 tr\u1ecb \u0111\u1ec1u th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n c\u1ee7a $x$<br\/> V\u1eady kh\u1ed1i l\u01b0\u1ee3ng h\u1ee3p kim l\u00fac \u0111\u1ea7u l\u00e0 $10$$kg$ ho\u1eb7c $25$$kg$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 A v\u00e0 D <\/span><\/span>"}]}],"id_ques":1016},{"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":"<span class='basic_left'>M\u1ed9t m\u1ea3nh \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chi\u1ec1u d\u00e0i $30$ $m$ v\u00e0 chi\u1ec1u r\u1ed9ng $20$ $m.$ \u1ede chung quanh m\u1ea3nh \u0111\u1ea5t v\u1ec1 ph\u00eda trong m\u1ea3nh \u0111\u1ea5t, ng\u01b0\u1eddi ta \u0111\u1ec3 m\u1ed9t l\u1ed1i \u0111i c\u00f3 chi\u1ec1u r\u1ed9ng kh\u00f4ng \u0111\u1ed5i, ph\u1ea7n c\u00f2n l\u1ea1i l\u00e0 m\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt \u0111\u01b0\u1ee3c tr\u1ed3ng hoa. Bi\u1ebft r\u1eb1ng di\u1ec7n t\u00edch v\u01b0\u1eddn hoa b\u1eb1ng $84$$\\%$ di\u1ec7n t\u00edch m\u1ea3nh \u0111\u1ea5t. Chi\u1ec1u r\u1ed9ng c\u1ee7a l\u1ed1i \u0111i l\u00e0:","select":["A. $1$ $m$ ","B. $1,5$ $m$","C. $2$ $m$ ","D. $0,5$ $m$ "],"hint":"V\u1ebd h\u00ecnh r\u1ed3i bi\u1ec3u di\u1ec5n chi\u1ec1u d\u00e0i v\u00e0 chi\u1ec1u r\u1ed9ng c\u1ee7a v\u01b0\u1eddn hoa qua chi\u1ec1u r\u1ed9ng c\u1ee7a l\u1ed1i \u0111i.","explain":"<span class='basic_left'><center><img src= 'https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai23/lv3/img\/K23.1.png' \/><\/center>G\u1ecdi chi\u1ec1u r\u1ed9ng c\u1ee7a l\u1ed1i \u0111i l\u00e0 $x$ $(m).$ <br\/>Chi\u1ec1u d\u00e0i c\u1ee7a v\u01b0\u1eddn hoa l\u00e0 $30-2x$ $(m),$ chi\u1ec1u r\u1ed9ng c\u1ee7a v\u01b0\u1eddn hoa l\u00e0 $20-2x$ $(m).$ \u0110i\u1ec1u ki\u1ec7n: $ 0 < x < 10 $<br\/>Di\u1ec7n t\u00edch v\u01b0\u1eddn hoa l\u00e0: $\\left( 30-2x \\right)\\left( 20-2x \\right)\\,\\,\\left( {{m}^{2}} \\right)$ <br\/>Di\u1ec7n t\u00edch m\u1ea3nh \u0111\u1ea5t l\u00e0: $20.30=600\\,\\,\\left( {{m}^{2}} \\right)$ <br\/>V\u00ec di\u1ec7n t\u00edch v\u01b0\u1eddn hoa b\u1eb1ng $84\\%$ di\u1ec7n t\u00edch m\u1ea3nh \u0111\u1ea5t n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: <br\/>$\\begin{aligned} & \\,\\,\\,\\,\\,\\,\\left( 30-2x \\right)\\left( 20-2x \\right)=84\\%.600 \\\\ & \\Leftrightarrow 600+4{{x}^{2}}-40x-60x=504 \\\\ & \\Leftrightarrow {{x}^{2}}-25x+24=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=1 \\\\ & x=24 \\\\ \\end{aligned} \\right.\\,\\left(\\text{do}\\, a+b+c=0 \\right) \\\\ & \\Leftrightarrow x=1\\,\\,\\left(\\text{do}\\, 0 < x < 10 \\right) \\\\ \\end{aligned}$<br\/>V\u1eady chi\u1ec1u r\u1ed9ng c\u1ee7a l\u1ed1i \u0111i l\u00e0 $1 m.$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><\/span>","column":4}]}],"id_ques":1017},{"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'>Ba c\u00f4ng nh\u00e2n c\u00f9ng l\u00e0m m\u1ed9t c\u00f4ng vi\u1ec7c th\u00ec l\u00e0m xong s\u1edbm h\u01a1n $18$ gi\u1edd so v\u1edbi khi ng\u01b0\u1eddi th\u1ee9 ba l\u00e0m m\u1ed9t m\u00ecnh; s\u1edbm h\u01a1n $3$ gi\u1edd so v\u1edbi khi ng\u01b0\u1eddi hai l\u00e0m m\u1ed9t m\u00ecnh v\u00e0 b\u1eb1ng n\u1eeda th\u1eddi gian so v\u1edbi khi ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t l\u00e0m m\u1ed9t m\u00ecnh. Th\u1eddi gian c\u00f4ng nh\u00e2n th\u1ee9 nh\u1ea5t, hai v\u00e0 ba l\u00e0m m\u1ed9t m\u00ecnh xong c\u00f4ng vi\u1ec7c l\u1ea7n l\u01b0\u1ee3t l\u00e0:","select":["A. $5$ gi\u1edd; $6$ gi\u1edd v\u00e0 $20 $ gi\u1edd","B. $4$ gi\u1edd; $6$ gi\u1edd v\u00e0 $20 $ gi\u1edd","C. $4$ gi\u1edd; $5$ gi\u1edd v\u00e0 $20 $ gi\u1edd ","D. $4$ gi\u1edd; $5$ gi\u1edd v\u00e0 $16$ gi\u1edd"],"hint":"G\u1ecdi th\u1eddi gian ng\u01b0\u1eddi I,II,III l\u00e0m m\u1ed9t m\u00ecnh xong c\u00f4ng vi\u1ec7c theo th\u1ee9 t\u1ef1 l\u00e0 $x;y;z$<br\/>Ch\u00fa \u00fd: S\u1ed1 ph\u1ea7n c\u00f4ng vi\u1ec7c m\u00e0 m\u1ed7i ng\u01b0\u1eddi l\u00e0m \u0111\u01b0\u1ee3c trong m\u1ed9t gi\u1edd v\u00e0 s\u1ed1 gi\u1edd c\u1ea7n thi\u1ebft \u0111\u1ec3 ng\u01b0\u1eddi \u0111\u00f3 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 th\u1eddi gian ng\u01b0\u1eddi I,II,III l\u00e0m m\u1ed9t m\u00ecnh xong c\u00f4ng vi\u1ec7c theo th\u1ee9 t\u1ef1 l\u00e0 $x;y;z$ $(x,y,z>0).$<br\/>Trong m\u1ed9t gi\u1edd, m\u1ed7i ng\u01b0\u1eddi I,II,II l\u00e0m \u0111\u01b0\u1ee3c l\u1ea7n l\u01b0\u1ee3t $\\dfrac{1}{x};\\dfrac{1}{y};\\dfrac{1}{z}$ (c\u00f4ng vi\u1ec7c), c\u1ea3 ba ng\u01b0\u1eddi l\u00e0m \u0111\u01b0\u1ee3c: $\\dfrac{1}{x}+\\dfrac{1}{y}+\\dfrac{1}{z}$ (c\u00f4ng vi\u1ec7c).<br\/>Th\u1eddi gian c\u1ea3 ba ng\u01b0\u1eddi c\u00f9ng l\u00e0m xong c\u00f4ng vi\u1ec7c l\u00e0 $\\dfrac{1}{\\dfrac{1}{x}+\\dfrac{1}{y}+\\dfrac{1}{z}}$ <br\/>Theo gi\u1ea3 thi\u1ebft, ta c\u00f3:<br\/>$\\begin{aligned} & \\,\\,\\,\\,\\,\\,\\dfrac{1}{\\dfrac{1}{x}+\\dfrac{1}{y}+\\dfrac{1}{z}}=z-18=y-3=\\dfrac{1}{2}x \\\\ & \\Rightarrow \\dfrac{2}{x}=\\dfrac{1}{z-18}=\\dfrac{1}{y-3}=\\dfrac{1}{x}+\\dfrac{1}{y}+\\dfrac{1}{z} \\\\ \\end{aligned}$ <br\/>Bi\u1ec3u th\u1ecb $z$ theo $ x$ \u0111\u01b0\u1ee3c: $z=\\dfrac{x+36}{2}$ (1)<br\/>Bi\u1ec3u th\u1ecb $y$ theo $x$ \u0111\u01b0\u1ee3c: $y=\\dfrac{x+6}{2}$ (2)<br\/>Thay c\u00e1c bi\u1ec3u di\u1ec5n tr\u00ean v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh: $\\dfrac{1}{x}+\\dfrac{1}{y}+\\dfrac{1}{z}=\\dfrac{2}{x}$ , ta \u0111\u01b0\u1ee3c:<br\/>$\\begin{aligned} & \\,\\,\\,\\,\\,\\,\\dfrac{1}{x}+\\dfrac{2}{x+6}+\\dfrac{2}{x+36}=\\dfrac{2}{x} \\\\ & \\Rightarrow 2x\\left( x+36 \\right)+2x\\left( x+6 \\right)-\\left( x+6 \\right)\\left( x+36 \\right)=0 \\\\ & \\Leftrightarrow 2{{x}^{2}}+72x+2{{x}^{2}}+12x-{{x}^{2}}-42x-216=0 \\\\ & \\Leftrightarrow 3{{x}^{2}}+42x-216=0 \\\\ & \\Leftrightarrow {{x}^{2}}+14x-72=0 \\\\ \\end{aligned}$<br\/>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c: $x_1=4;x_2=-18 $(lo\u1ea1i). V\u1eady $ x=4$<br\/>Thay $x=4$ v\u00e0o (1) v\u00e0 (2), ta \u0111\u01b0\u1ee3c: $z=20;y=5$<br\/>V\u1eady th\u1eddi gian ng\u01b0\u1eddi I,II,III l\u00e0m m\u1ed9t m\u00ecnh xong c\u00f4ng vi\u1ec7c theo th\u1ee9 t\u1ef1 l\u00e0 $4$ gi\u1edd; $5$ gi\u1edd v\u00e0 $20 $ gi\u1edd.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":2}]}],"id_ques":1018},{"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":"<span class='basic_left'>M\u1ed9t c\u1eeda h\u00e0ng mua $x$ chi\u1ebfc \u00e1o h\u1ebft $d$ ngh\u00ecn \u0111\u1ed3ng. C\u1eeda h\u00e0ng b\u00e1n $2$ chi\u1ebfc v\u1edbi gi\u00e1 b\u1eb1ng m\u1ed9t n\u1eeda gi\u00e1 mua, b\u00e1n nh\u1eefng chi\u1ebfc c\u00f2n l\u1ea1i \u0111\u01b0\u1ee3c l\u00e3i $8$ ngh\u00ecn \u0111\u1ed3ng m\u1ed9t chi\u1ebfc. Ti\u1ec1n l\u00e3i t\u1ed5ng c\u1ed9ng l\u00e0 $72$ ngh\u00ecn \u0111\u1ed3ng. T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $x$ bi\u1ebft r\u1eb1ng $d$ l\u00e0 m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean.","select":["A. $x=11$","B. $x=12$","C. $x=2 $","D. $x=10$"],"hint":"L\u1eadp ph\u01b0\u01a1ng tr\u00ecnh bi\u1ec3u th\u1ecb s\u1ed1 ti\u1ec1n l\u00e3i thu \u0111\u01b0\u1ee3c.","explain":"<span class='basic_left'>Gi\u00e1 mua m\u1ed9t chi\u1ebfc \u00e1o l\u00e0 $\\dfrac{d}{x}$ (ngh\u00ecn \u0111\u1ed3ng). <br\/>S\u1ed1 ti\u1ec1n b\u00e1n $2$ chi\u1ebfc v\u1edbi gi\u00e1 b\u1eb1ng m\u1ed9t n\u1eeda gi\u00e1 mua l\u00e0 $2.\\dfrac{d}{2x}=\\dfrac{d}{x}$ (ngh\u00ecn \u0111\u1ed3ng). <br\/>Gi\u00e1 ti\u1ec1n b\u00e1n nh\u1eefng chi\u1ebfc c\u00f2n l\u1ea1i (l\u00e3i $8$ ngh\u00ecn \u0111\u1ed3ng m\u1ed9t chi\u1ebfc) l\u00e0: $\\left( x-2 \\right)\\left( \\dfrac{d}{x}+8 \\right)$ (ngh\u00ecn \u0111\u1ed3ng).<br\/>Ti\u1ec1n l\u00e3i t\u1ed5ng c\u1ed9ng l\u00e0 $72$ ngh\u00ecn \u0111\u1ed3ng n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: <br\/>$\\begin{align} & \\,\\,\\,\\,\\,\\,\\dfrac{d}{x}+\\left( x-2 \\right)\\left( \\dfrac{d}{x}+8 \\right)=72 \\\\ & \\Leftrightarrow \\,\\dfrac{d}{x}+d-\\dfrac{2d}{x}+8x-16=72 \\\\ & \\Leftrightarrow 8x-\\dfrac{d}{x}-88=0 \\\\ & \\Rightarrow 8{{x}^{2}}-88x=d \\\\ & \\Leftrightarrow 8x\\left( x-11 \\right)=d \\\\ \\end{align}$ <br\/>Do $d$ v\u00e0 $x$ l\u00e0 s\u1ed1 nguy\u00ean d\u01b0\u01a1ng n\u00ean $x-11$ nguy\u00ean d\u01b0\u01a1ng. Suy ra $x>11$<br\/>Gi\u00e1 tr\u1ecb nguy\u00ean d\u01b0\u01a1ng nh\u1ecf nh\u1ea5t c\u1ee7a $x$ l\u00e0 $12.$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":1019},{"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":[[["40"],["60"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>Hai b\u00e0 ra ch\u1ee3 b\u00e1n t\u1ed5ng c\u1ed9ng $100$ qu\u1ea3 tr\u1ee9ng. S\u1ed1 tr\u1ee9ng c\u1ee7a hai ng\u01b0\u1eddi b\u00e1n kh\u00f4ng b\u1eb1ng nhau nh\u01b0ng s\u1ed1 ti\u1ec1n thu \u0111\u01b0\u1ee3c l\u1ea1i b\u1eb1ng nhau.<br\/>B\u00e0 th\u1ee9 nh\u1ea5t n\u00f3i v\u1edbi b\u00e0 th\u1ee9 hai:<br\/>- N\u1ebfu t\u00f4i c\u00f3 s\u1ed1 tr\u1ee9ng nh\u01b0 c\u1ee7a b\u00e0, t\u00f4i s\u1ebd thu \u0111\u01b0\u1ee3c $15$ \u0111\u1ed3ng.<br\/>B\u00e0 th\u1ee9 2 n\u00f3i:<br\/>- N\u1ebfu s\u1ed1 tr\u1ee9ng c\u1ee7a t\u00f4i b\u1eb1ng s\u1ed1 tr\u1ee9ng c\u1ee7a b\u00e0, t\u00f4i ch\u1ec9 b\u00e1n \u0111\u01b0\u1ee3c $6\\dfrac{2}{3}$ \u0111\u1ed3ng.<br\/>H\u1ecfi m\u1ed7i b\u00e0 c\u00f3 bao nhi\u00eau qu\u1ea3 tr\u1ee9ng mang \u0111i b\u00e1n?<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> B\u00e0 th\u1ee9 nh\u1ea5t: _input_ (qu\u1ea3); b\u00e0 th\u1ee9 hai: _input_(qu\u1ea3)<\/span>","hint":"D\u1ef1a v\u00e0o l\u1eddi n\u00f3i c\u1ee7a m\u1ed7i b\u00e0, ta t\u00ednh gi\u00e1 ti\u1ec1n m\u1ed9t qu\u1ea3 tr\u1ee9ng c\u1ee7a t\u1eebng b\u00e0","explain":"<span class='basic_left'><span class='basic_green'>Ph\u00e2n t\u00edch b\u00e0i to\u00e1n:<\/span><br\/> <table><tr><th>C\u00e1c \u0111\u1ea1i l\u01b0\u1ee3ng<br><\/th><th>S\u1ed1 tr\u1ee9ng (qu\u1ea3)<br><\/th><th>S\u1ed1 ti\u1ec1n m\u1ed7i qu\u1ea3 tr\u1ee9ng (\u0111\u1ed3ng)<br><\/th><th>S\u1ed1 ti\u1ec1n b\u00e1n \u0111\u01b0\u1ee3c (\u0111\u1ed3ng)<br><\/th><\/tr><tr><th>B\u00e0 th\u1ee9 nh\u1ea5t<br><\/th><td>$x$<\/td><td>$\\dfrac{15}{100-x}$<\/td><td>$\\dfrac{15}{100-x}.x$<\/td><\/tr><tr><th>B\u00e0 th\u1ee9 hai<br><\/th><td>$100-x$<\/td><td>$6\\dfrac{2}{3}:x=\\dfrac{20}{3x}$<\/td><td>$\\dfrac{20}{3x}.\\left( 100-x \\right)$<\/td><\/tr><\/table><br\/>C\u0103n c\u1ee9 v\u00e0o s\u1ed1 ti\u1ec1n m\u1ed7i ng\u01b0\u1eddi thu \u0111\u01b0\u1ee3c l\u00e0 b\u1eb1ng nhau, ta l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh.<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>G\u1ecdi s\u1ed1 tr\u1ee9ng c\u1ee7a b\u00e0 th\u1ee9 nh\u1ea5t mang \u0111i b\u00e1n l\u00e0 $x$ (qu\u1ea3), $x\\in {{\\mathbb{N}}^{*}},x<100$ <br\/>Khi \u0111\u00f3 s\u1ed1 tr\u1ee9ng b\u00e0 th\u1ee9 hai mang \u0111i b\u00e1n l\u00e0 $100-x$ (qu\u1ea3)<br\/>Theo l\u1eddi n\u00f3i c\u1ee7a b\u00e0 th\u1ee9 nh\u1ea5t th\u00ec gi\u00e1 ti\u1ec1n m\u1ed7i qu\u1ea3 tr\u1ee9ng b\u00e0 th\u1ee9 nh\u1ea5t b\u00e1n l\u00e0 $\\dfrac{15}{100-x}$ (\u0111\u1ed3ng).<br\/>Theo l\u1eddi n\u00f3i c\u1ee7a b\u00e0 th\u1ee9 hai, gi\u00e1 ti\u1ec1n m\u1ed7i qu\u1ea3 tr\u1ee9ng c\u1ee7a b\u00e0 th\u1ee9 hai l\u00e0 $6\\dfrac{2}{3}:x=\\dfrac{20}{3x}$ (\u0111\u1ed3ng) <br\/>Do s\u1ed1 ti\u1ec1n m\u1ed7i ng\u01b0\u1eddi thu \u0111\u01b0\u1ee3c l\u00e0 b\u1eb1ng nhau n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\dfrac{15}{100-x}.x=\\dfrac{20}{3x}.\\left( 100-x \\right)$ <br\/>Thu g\u1ecdn ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c: ${{x}^{2}}+160x-8000=0$ <br\/>$\\begin{align} & \\Delta '=14400\\Rightarrow \\sqrt{\\Delta '}=120 \\\\ & \\Rightarrow {{x}_{1}}=-80+120=40;{{x}_{2}}=-80-120=-200 \\\\ \\end{align}$<br\/>V\u00ec $x\\in {{\\mathbb{N}}^{*}},x<100$ n\u00ean $x=x_1=40$<br\/>V\u1eady b\u00e0 th\u1ee9 nh\u1ea5t c\u00f3 $40$ qu\u1ea3 tr\u1ee9ng, b\u00e0 th\u1ee9 hai c\u00f3 $60$ qu\u1ea3 tr\u1ee9ng<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $40$ v\u00e0 $60$ <\/span><\/span>"}]}],"id_ques":1020}],"lesson":{"save":0,"level":3}}