{"segment":[{"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai23/lv3/img\/4.jpg' \/><\/center>Trong c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh sau, ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t 1 \u1ea9n l\u00e0:","select":["A. $x^2+2x+1=0$","B. $x+2y=3$","C. $\\dfrac{2}{x}=x+3$","D. $\\dfrac{1}{3}y=3$"],"hint":"","explain":" <span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 d\u1ea1ng $ax+b=0$ (trong \u0111\u00f3, $a$ v\u00e0 $b$ l\u00e0 c\u00e1c s\u1ed1 cho tr\u01b0\u1edbc $a\\ne 0$)<br\/>$\\bullet \\,\\, x^2+2x+1=0$ l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai 2 \u1ea9n<br\/>$\\bullet\\,\\,x+2y=3$ l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t 2 \u1ea9n <br\/>$\\bullet\\,\\,\\dfrac{2}{x}=x+3 \\,\\Rightarrow x^2+3x-2=0$ l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai m\u1ed9t \u1ea9n<br\/>$\\bullet\\,\\, \\dfrac{1}{3}y=3$ l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n. <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><\/span>","column":2}],"id_ques":1041},{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. {$\\dfrac{13}{5};\\dfrac{17}{5}$}","B. {$\\dfrac{3}{5};\\dfrac{17}{5}$}","C. {$\\dfrac{3}{5};\\dfrac{7}{5}$}"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai23/lv3/img\/16.jpg' \/><\/center>Ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{1}{4}(x^2-6x+9)-\\dfrac{1}{25}=0$ c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0: $S=?$","hint":"S\u1eed d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $a^2-b^2$, ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\dfrac{1}{4}(x^2-6x+9)-\\dfrac{1}{25}=0\\\\ \\Leftrightarrow \\left(\\dfrac{x-3}{2}\\right)^2-\\left(\\dfrac{1}{5}\\right)^2=0\\\\ \\Leftrightarrow \\left(\\dfrac{x-3}{2}-\\dfrac{1}{5}\\right).\\left(\\dfrac{x-3}{2}+\\dfrac{1}{5}\\right)=0\\\\ \\Leftrightarrow \\left[\\begin{aligned}&\\dfrac{x-3}{2}-\\dfrac{1}{5}=0\\\\ &\\dfrac{x-3}{2}+\\dfrac{1}{5}=0\\\\ \\end{aligned}\\right.\\\\ \\Leftrightarrow \\left[\\begin{aligned}&\\dfrac{x-3}{2}=\\dfrac{1}{5}\\\\ &\\dfrac{x-3}{2}=-\\dfrac{1}{5}\\\\ \\end{aligned}\\right.\\\\ \\Leftrightarrow \\left[\\begin{aligned}&5(x-3)=2\\\\ &5(x-3)=-2\\\\ \\end{aligned}\\right.\\\\ \\Leftrightarrow \\left[\\begin{aligned}&5x=17\\\\ &5x=13\\\\ \\end{aligned}\\right.\\\\ \\Leftrightarrow\\left[\\begin{aligned}&x=\\dfrac{17}{5}\\\\ &x=\\dfrac{13}{5}\\\\ \\end{aligned}\\right.$<\/span>"}],"id_ques":1042},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai23/lv3/img\/9.jpg' \/><\/center>Ph\u01b0\u01a1ng tr\u00ecnh $2x+4=0$ t\u01b0\u01a1ng \u0111\u01b0\u01a1ng v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh n\u00e0o sau \u0111\u00e2y.","select":["A. $x^2-x-6=0$","B. $\\dfrac{1}{2}x+1=0$","C. $x^2-4=3$","D. $3x-6=0$"],"hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh \u1edf m\u1ed7i \u0111\u00e1p \u00e1n v\u00e0 so s\u00e1nh v\u1edbi t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u1edf \u0111\u1ea7u b\u00e0i.","explain":" <span class='basic_left'>$2x+4=0\\Leftrightarrow x=-2$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh $2x+4=0$ c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{-2\\}$<br\/>A. $x^2-x-6=0\\\\ \\Leftrightarrow x^2-3x+2x-6=0\\\\ \\Leftrightarrow (x-3)(x+2)=0\\\\ \\Leftrightarrow \\left[\\begin{aligned}&x=3\\\\ &x=-2\\\\ \\end{aligned}\\right.$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{-2;3\\}$<br\/>B. $\\dfrac{1}{2}x+1=0\\\\ \\Leftrightarrow x=-2$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{-2\\}$<br\/>C. $x^2-4=5\\\\ \\Leftrightarrow x^2=9\\\\ \\Leftrightarrow \\left[\\begin{aligned}&x=-3\\\\ &x=3\\\\ \\end{aligned}\\right.$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{-3;3\\}$<br\/>D. $3x-6=0\\Leftrightarrow x=2$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{2\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}],"id_ques":1043},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai23/lv3/img\/3.jpg' \/><\/center>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x-1}{x-2}-\\dfrac{2}{x-3}=\\dfrac{x}{x^2-5x+6}.$","select":["A. $x\\ne 2$","B. $x\\ne 3$","C. $x\\ne 2$ ho\u1eb7c $ x\\ne 3$ ","D. $x\\ne 2$ v\u00e0 $x\\ne 3$"],"hint":"","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x-1}{x-2}-\\dfrac{2}{x-3}=\\dfrac{x}{x^2-5x+6}$ l\u00e0: <br\/>$\\left \\{\\begin {aligned}&x-2\\ne 0\\\\ &x-3\\ne 0\\\\ & x^2-5x+6\\ne 0\\\\ \\end{aligned}\\right.\\Leftrightarrow \\left \\{\\begin {aligned}&x\\ne 2\\\\ &x\\ne 3\\\\ & (x-2)(x-3)\\ne 0\\\\ \\end{aligned}\\right.\\Leftrightarrow\\left \\{\\begin {aligned}&x\\ne 2\\\\ &x\\ne 3\\\\ \\end{aligned}\\right.$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><\/span>","column":2}],"id_ques":1044},{"title":"L\u1ef1a ch\u1ecdn \u0110\u00fang ho\u1eb7c Sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai23/lv3/img\/2.jpg' \/><\/center>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $(x+3)^2-(x-3)^2=6x+18$ l\u00e0 $x=3.$","select":["\u0110\u00fang","Sai"],"hint":"","explain":" <span class='basic_left'> Ta c\u00f3:<br\/>$(x+3)^2-(x-3)^2=6x+18\\\\ \\Leftrightarrow (x^2+6x+9)-(x^2-6x+9)=6x+18\\\\ \\Leftrightarrow x^2+6x+9-x^2+6x-9-6x-18=0\\\\ \\Leftrightarrow 6x=18\\\\ \\Leftrightarrow x=3$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{3\\}$<br\/><span class='basic_pink'>Kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 Sai<\/span><br\/><span class='basic_green'>Ghi nh\u1edb: <\/span>T\u1eadp h\u1ee3p nghi\u1ec7m c\u00f3 d\u1ea1ng $S=\\{x_1,; x_2;x_3...\\}$, trong \u0111\u00f3 $x_1; x_2; x_3.. $ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh.<\/span>","column":2}],"id_ques":1045},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["2"],["3"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai23/lv3/img\/16.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $x(x-2)-3x+6=0.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${_input_;_input_}","hint":"\u0110\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh t\u00edch","explain":"<span class='basic_left'>Ta c\u00f3: <br\/>$x(x-2)-3x+6=0\\\\ \\Leftrightarrow x(x-2)-3(x-2)=0\\\\ \\Leftrightarrow (x-2)(x-3)=0\\\\ \\Leftrightarrow \\left[\\begin{aligned}&x=2\\\\ &x=3\\\\ \\end{aligned}\\right.$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{2;3\\}$<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2$ v\u00e0 $3$<\/span><\/span>"}],"id_ques":1046},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai23/lv3/img\/13.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $3+\\dfrac{3+x}{3-x}-\\dfrac{3x-1}{x+3}=\\dfrac{35}{{{x}^{2}}-9}.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","hint":"","select":["A. $S=\\{3;-19\\}$","B. $S=\\{-2,9;-18,9\\}$","C. $S=\\mathbb R$ ","D. $S=\\varnothing$"],"explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 3$ v\u00e0 $x\\ne -3$<br\/>$\\begin{aligned} & 3+\\dfrac{3+x}{3-x}-\\dfrac{3x-1}{x+3}=\\dfrac{35}{{{x}^{2}}-9} \\\\ & \\Leftrightarrow \\dfrac{3({{x}^{2}}-9)}{(x-3)(x+3)}-\\dfrac{(x+3)(x+3)}{(x-3)(x+3)}-\\dfrac{(3x-1)(x-3)}{(x+3)(x-3)}=\\dfrac{35}{(x-3)(x+3)} \\\\ & \\Rightarrow 3({{x}^{2}}-9)-{{(x+3)}^{2}}-(3x-1)(x-3)=35 \\\\ & \\Leftrightarrow 3{{x}^{2}}-27-{{x}^{2}}-6x-9-(3{{x}^{2}}-10x+3)-35=0 \\\\ & \\Leftrightarrow -{{x}^{2}}+4x-74=0 \\\\ & \\Leftrightarrow {{x}^{2}}-4x+4+70=0 \\\\ & \\Leftrightarrow {{(x-2)}^{2}}+70=0 \\\\ & \\Leftrightarrow {{(x-2)}^{2}}=-70\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{v\u00f4 nghi\u1ec7m}) \\\\ \\end{aligned}$ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho v\u00f4 nghi\u1ec7m<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><\/span>","column":2}],"id_ques":1047},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["60"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai23/lv3/img\/9.jpg' \/><\/center><span class='basic_left'>L\u00fac $8$ gi\u1edd, m\u1ed9t ng\u01b0\u1eddi \u0111i xe m\u00e1y t\u1eeb $A$ \u0111\u1ebfn $B$ v\u1edbi v\u1eadn t\u1ed1c $40 km\/h$. \u0110\u1ebfn $B$ ng\u01b0\u1eddi \u0111\u00f3 l\u00e0m vi\u1ec7c trong $3$ gi\u1edd r\u1ed3i quay v\u1ec1 $A$ v\u1edbi v\u1eadn t\u1ed1c $30 km\/h$. T\u00ednh \u0111\u1ed9 d\u00e0i qu\u00e3ng \u0111\u01b0\u1eddng $AB$. Bi\u1ebft ng\u01b0\u1eddi \u0111\u00f3 v\u1ec1 \u0111\u1ebfn $A$ l\u00fac $14$ gi\u1edd $30$ ph\u00fat.<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> \u0110\u1ed9 d\u00e0i qu\u00e3ng \u0111\u01b0\u1eddng $AB$ l\u00e0 _input_ ($km$)<\/span>","hint":"S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i b\u00e0i to\u00e1n chuy\u1ec3n \u0111\u1ed9ng","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/>L\u1eadp b\u1ea3ng:<br\/><table><tr><th><\/th><th>Qu\u00e3ng \u0111\u01b0\u1eddng ($km$)<\/th><th>V\u1eadn t\u1ed1c ($km\/h$)<\/th><th>Th\u1eddi gian (gi\u1edd)<\/th><\/tr><tr><td>\u0110i t\u1eeb $A$ \u0111\u1ebfn $B$<\/td><td>$x$<\/td><td>$40$<\/td><td>$\\dfrac {x}{40}$<\/td><\/tr><tr><td>L\u00e0m vi\u1ec7c<\/td><td>$0$<\/td><td>$0$<\/td><td>$3$<\/td><\/tr><tr><td>\u0110i t\u1eeb $B$ v\u1ec1 $A$<\/td><td>$x$<\/td><td>$30$<\/td><td>$\\dfrac {x}{30}$<\/td><\/tr><\/table><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>G\u1ecdi qu\u00e3ng \u0111\u01b0\u1eddng $AB$ d\u00e0i $x$ ($km,\\,x>0$)<br\/>Th\u1eddi gian ng\u01b0\u1eddi \u0111\u00f3 \u0111i t\u1eeb $A$ \u0111\u1ebfn $B$ l\u00e0 $\\dfrac {x}{40}$ (gi\u1edd)<br\/>Ng\u01b0\u1eddi \u0111\u00f3 l\u00e0m vi\u1ec7c t\u1ea1i $B$ trong $3$ gi\u1edd.<br\/>Th\u1eddi gian ng\u01b0\u1eddi \u0111\u00f3 \u0111i t\u1eeb $B$ v\u1ec1 $A$ l\u00e0 $\\dfrac{x}{30}$(gi\u1edd)<br\/>T\u1ed5ng th\u1eddi gian \u0111i, v\u1ec1 v\u00e0 l\u00e0m vi\u1ec7c t\u1ea1i $B$ l\u00e0 $6$ gi\u1edd $30$ ph\u00fat $= \\dfrac {13}{2}$ (gi\u1edd)<br\/>Ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: <br\/>$\\begin{aligned}& \\dfrac{x}{40}+3+\\dfrac{x}{30}=\\dfrac{13}{2} \\\\ & \\Leftrightarrow 3x+360+4x=780 \\\\ & \\Leftrightarrow 7x=420 \\\\ & \\Leftrightarrow x=60\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ \\end{aligned}$ <br\/>V\u1eady \u0111\u1ed9 d\u00e0i qu\u00e3ng \u0111\u01b0\u1eddng $AB$ l\u00e0 $60\\,km$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $60$<\/span><\/span>"}],"id_ques":1048},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["5"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai23/lv3/img\/10.jpg' \/><\/center>Hai v\u00f2i n\u01b0\u1edbc c\u00f9ng ch\u1ea3y v\u00e0o b\u1ec3 sau $2$ gi\u1edd th\u00ec \u0111\u1ea7y b\u1ec3. M\u1ed7i gi\u1edd l\u01b0\u1ee3ng n\u01b0\u1edbc v\u00f2i $1$ ch\u1ea3y v\u00e0o b\u1ec3 b\u1eb1ng $\\dfrac {2}{3}$ l\u01b0\u1ee3ng n\u01b0\u1edbc v\u00f2i $2$ ch\u1ea3y v\u00e0o b\u1ec3. T\u00ednh th\u1eddi gian v\u00f2i $1$ ch\u1ea3y m\u1ed9t m\u00ecnh \u0111\u1ea7y b\u1ec3.<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> _input_ (gi\u1edd)","hint":"S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i b\u00e0i to\u00e1n n\u0103ng su\u1ea5t l\u00e0m chung - l\u00e0m ri\u00eang","explain":"<span class='basic_left'>G\u1ecdi th\u1eddi gian v\u00f2i $1$ ch\u1ea3y m\u1ed9t m\u00ecnh \u0111\u1ea7y b\u1ec3 l\u00e0 $x$ (gi\u1edd, $x>2$)<br\/>M\u1ed9t gi\u1edd v\u1eddi $1$ ch\u1ea3y \u0111\u01b0\u1ee3c l\u00e0 $\\dfrac {1}{x}$ (b\u1ec3)<br\/>V\u00ec hai v\u00f2i ch\u1ea3y trong $2$ gi\u1edd th\u00ec \u0111\u1ea7y b\u1ec3 n\u00ean $1$ gi\u1edd c\u1ea3 hai v\u00f2i ch\u1ea3y \u0111\u01b0\u1ee3c l\u00e0 $\\dfrac{1}{2}$ (b\u1ec3)<br\/>V\u00ec $1$ gi\u1edd v\u00f2i $1$ ch\u1ea3y \u0111\u01b0\u1ee3c b\u1eb1ng $\\dfrac{2}{3}$ v\u00f2i $2$ n\u00ean $1$ gi\u1edd v\u00f2i $2$ ch\u1ea3y \u0111\u01b0\u1ee3c l\u00e0 $\\dfrac {3}{2}.\\dfrac{1}{x}=\\dfrac{3}{2x}$ (b\u1ec3)<br\/>Ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\begin{aligned} & \\dfrac{1}{x}+\\dfrac{3}{2x}=\\dfrac{1}{2} \\\\ & \\Leftrightarrow \\dfrac{2}{2x}+\\dfrac{3}{2x}=\\dfrac{x}{2x}\\\\ & \\Rightarrow 2+3=x \\\\ & \\Leftrightarrow x=5\\,\\,\\text{(th\u1ecfa m\u00e3n)} \\\\ \\end{aligned}$<br\/>V\u1eady v\u00f2i $1$ ch\u1ea3y m\u1ed9t m\u00ecnh trong $5$ gi\u1edd th\u00ec \u0111\u1ea7y b\u1ec3<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $5$<\/span> <\/span>"}],"id_ques":1049},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai23/lv3/img\/12.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $(x^2+6x+8)(x^2+8x+15)-24=0.$","select":["A. $S=\\{-1;-6\\}$","B. $S=\\{-6;4\\}$","C. $S=\\mathbb R$","D. $S=\\varnothing$"],"hint":"Ph\u00e2n t\u00edch bi\u1ec3u th\u1ee9c trong ngo\u1eb7c th\u00e0nh nh\u00e2n t\u1eed. S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p nh\u00f3m r\u1ed3i th\u1ef1c hi\u1ec7n ph\u00e9p nh\u00e2n \u0111\u1ec3 t\u00ecm \u1ea9n ph\u1ee5.<br\/>\u0110\u1eb7t \u1ea9n ph\u1ee5.","explain":" <span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned} & ({{x}^{2}}+6x+8)({{x}^{2}}+8x+15)-24=0 \\\\ & \\Leftrightarrow (x^2+2x+4x+8)(x^2+3x+5x+15)-24=0\\\\ & \\Leftrightarrow [x(x+2)+4(x+2)].[x(x+3)+5(x+3)]-24=0\\\\ & \\Leftrightarrow (x+2)(x+4)(x+3)(x+5)-24=0 \\\\ & \\Leftrightarrow [(x+2)(x+5)][(x+3)(x+4)]-24=0 \\\\ & \\Leftrightarrow \\left( {{x}^{2}}+7x+10 \\right)\\left( {{x}^{2}}+7x+12 \\right)-24=0 \\\\ \\end{aligned}$ <br\/>\u0110\u1eb7t $x^2+7x+10=t$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh:<br\/>$\\begin{aligned} & t(t+2)-24=0 \\\\ & \\Leftrightarrow {{t}^{2}}+2t-24=0\\\\ &\\Leftrightarrow t^2-4t+6t-24=0\\\\ &\\Leftrightarrow t(t-4)+6(t-4)=0 \\\\ & \\Leftrightarrow (t-4)(t+6)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & t=4\\\\ & t=-6\\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/> V\u1edbi $t=4$ ta c\u00f3:<br\/>$\\begin{aligned} & {{x}^{2}}+7x+10=4 \\\\ & \\Leftrightarrow {{x}^{2}}+7x+6=0 \\\\ &\\Leftrightarrow x^2+x+6x+6=0\\\\ &\\Leftrightarrow x(x+1)+6(x+1)=0\\\\ & \\Leftrightarrow (x+1)(x+6)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=-1 \\\\ & x=-6 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/>V\u1edbi $t=-6$, ta c\u00f3:<br\/>$\\begin{aligned} & {{x}^{2}}+7x+10=-6 \\\\ & \\Leftrightarrow {{x}^{2}}+7x+16=0 \\\\ & \\Leftrightarrow {{x}^{2}}+2.x.\\dfrac{7}{2}+\\dfrac{49}{4}+\\dfrac{15}{4}=0 \\\\ & \\Leftrightarrow {{\\left( x+\\dfrac{7}{2} \\right)}^{2}}=\\dfrac{-15}{4}\\,\\,\\text{(v\u00f4 nghi\u1ec7m)} \\\\ \\end{aligned}$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{-1;-6\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><\/span>","column":2}],"id_ques":1050}],"id_ques":0}],"lesson":{"save":1,"level":3,"time":44}}