{"segment":[{"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'> Bi\u1ec3u th\u1ee9c n\u00e0o sau \u0111\u00e2y bi\u1ec3u th\u1ecb chu vi c\u1ee7a m\u1ed9t tam gi\u00e1c c\u00e2n c\u00f3 c\u1ea1nh b\u00ean b\u1eb1ng $a$ v\u00e0 c\u1ea1nh \u0111\u00e1y b\u1eb1ng $b.$ <\/span>","select":["A. $2a+b$","B. $2(a+b)$","C. $a+2b$ ","D. $2ab$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Chu vi tam gi\u00e1c b\u1ea5t k\u00ec b\u1eb1ng t\u1ed5ng \u0111\u1ed9 d\u00e0i ba c\u1ea1nh c\u1ee7a n\u00f3. <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Chu vi m\u1ed9t tam gi\u00e1c c\u00e2n c\u00f3 c\u1ea1nh b\u00ean b\u1eb1ng $a,$ c\u1ea1nh \u0111\u00e1y b\u1eb1ng $b$ vi\u1ebft th\u00e0nh bi\u1ec3u th\u1ee9c \u0111\u1ea1i s\u1ed1 l\u00e0: $a+a+b=2a+b$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span> <\/span>","column":2}]}],"id_ques":1091},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00e2y \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":" <span class='basic_left'> Bi\u1ec3u th\u1ee9c $(a-b)^2(a^2+b^2)$ di\u1ec5n \u0111\u1ea1t b\u1eb1ng l\u1eddi l\u00e0: <b> T\u00edch c\u1ee7a hi\u1ec7u $a$ v\u00e0 $b$ b\u00ecnh ph\u01b0\u01a1ng v\u1edbi t\u1ed5ng c\u00e1c b\u00ecnh ph\u01b0\u01a1ng c\u1ee7a $a$ v\u00e0 $b$ <\/b> <\/span>","select":["\u0110\u00fang","Sai "],"explain":"<span class='basic_left'> Bi\u1ec3u th\u1ee9c $(a-b)^2(a^2+b^2)$ di\u1ec5n \u0111\u1ea1t b\u1eb1ng l\u1eddi l\u00e0: <b> T\u00edch c\u1ee7a b\u00ecnh ph\u01b0\u01a1ng c\u1ee7a hi\u1ec7u $a$ v\u00e0 $b$ v\u1edbi t\u1ed5ng c\u00e1c b\u00ecnh ph\u01b0\u01a1ng c\u1ee7a $a$ v\u00e0 $b$ <\/b> <br\/> $\\Rightarrow$ Kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 sai. <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 Sai.<\/span><\/span>","column":2}]}],"id_ques":1092},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 \u0111\u00fang nh\u1ea5t v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["0"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $F=\\dfrac{3a-5}{2a+b}-\\dfrac{4b+5}{a+3b}$ v\u1edbi $a-b=5$ <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $F=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ <\/span>","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> T\u1eeb $a-b=5$ r\u00fat $a$ theo $b$ <br\/> \u0110\u01b0a $F$ v\u1ec1 bi\u1ec3u th\u1ee9c ch\u1ec9 ch\u1ee9a bi\u1ebfn $b$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> T\u1eeb $a-b=5 \\Rightarrow a=b+5$ <br\/> Khi \u0111\u00f3 ta c\u00f3: <br\/> $\\begin{align} & F=\\dfrac{3\\left( b+5 \\right)-5}{2(b+5)+b}-\\dfrac{4b+5}{(b+5)+3b} \\\\ & \\,\\,\\,\\,=\\dfrac{3b+15-5}{2b+10+b}-\\dfrac{4b+5}{b+5+3b} \\\\ & \\,\\,\\,\\,=\\dfrac{3b+10}{3b+10}-\\dfrac{4b+5}{4b+5} \\\\ & \\,\\,\\,\\,=1-1 \\\\ & \\,\\,\\,\\,=0 \\\\ \\end{align}$ <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":1093},{"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'> T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $D=x^2(x+y)-y^2(x+y)+x^2-y^2+2(x+y)+3$ bi\u1ebft r\u1eb1ng $x+y+1=0$ <\/span>","select":["A. $D=0$","B. $D=1$","C. $D=2$","D. $D=3$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> T\u1eeb $x+y+1=0$ r\u00fat ra gi\u00e1 tr\u1ecb c\u1ee7a $x+y$ <br\/> Thay gi\u00e1 tr\u1ecb c\u1ee7a t\u1ed5ng $x+y$ \u0111\u00f3 v\u00e0o $D$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> T\u1eeb $x+y+1=0 \\Rightarrow x+y=-1$ <br\/> Thay $x+y=-1$ v\u00e0o $D$ ta c\u00f3: <br\/> $\\begin{align} & D={{x}^{2}}.\\left( -1 \\right)-{{y}^{2}}.\\left( -1 \\right)+{{x}^{2}}-{{y}^{2}}+2.\\left( -1 \\right)+3 \\\\ & \\,\\,\\,\\,\\,=-{{x}^{2}}+{{y}^{2}}+{{x}^{2}}-{{y}^{2}}-2+3 \\\\ & \\,\\,\\,\\,\\,=1 \\\\ \\end{align}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span>","column":4}]}],"id_ques":1094},{"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'> T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $E=(x+y)(y+z)(x+z)$ bi\u1ebft r\u1eb1ng $x+y+z=0$ v\u00e0 $xyz=2$ <\/span>","select":["A. $E=0$","B. $E=1$","C. $E=-1$","D. $E=-2$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> T\u1eeb $x+y+z=0$ r\u00fat ra gi\u00e1 tr\u1ecb c\u1ee7a $x+y; \\,y+z$ v\u00e0 $x+z$ <br\/> Thay c\u00e1c t\u1ed5ng \u0111\u00f3 v\u00e0o $E$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> T\u1eeb $x+y+z=0 \\Rightarrow x+y=-z;\\, y+z=-x;\\, x+z=-y$ <br\/> Thay $x+y=-z;\\, y+z=-x;\\, x+z=-y$ v\u00e0o $E$ ta c\u00f3: <br\/> $E=\\left( -z \\right).\\left( -x \\right).\\left( -y \\right)=-xyz=-2$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span>","column":4}]}],"id_ques":1095},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 \u0111\u00fang nh\u1ea5t v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["-1"],["0"],["1"],["3"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> T\u00ecm gi\u00e1 tr\u1ecb nguy\u00ean c\u1ee7a bi\u1ebfn $x$ \u0111\u1ec3 bi\u1ec3u th\u1ee9c $N=\\dfrac{9x+5}{3x-1}$ nh\u1eadn gi\u00e1 tr\u1ecb nguy\u00ean. <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $\\left[ \\begin{align} & x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & x= \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\\\ & x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{align} \\right.$ <\/span>","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Ph\u00e2n t\u00edch bi\u1ec3u th\u1ee9c $N$ v\u1ec1 d\u1ea1ng $a+\\dfrac{b}{B(x)}$ v\u1edbi $a;b$ l\u00e0 h\u1eb1ng s\u1ed1, $B(x)$ l\u00e0 bi\u1ec3u th\u1ee9c ch\u1ee9a bi\u1ebfn $x$ <br\/> \u0110\u1ec3 $N\\in \\mathbb{Z}$ th\u00ec $B(x)\\in \u01af(b)$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3: <br\/> $N=\\dfrac{9x+5}{3x-1}=\\dfrac{3\\left( 3x-1 \\right)+8}{3x-1}=3+\\dfrac{8}{3x-1}$ <br\/> \u0110\u1ec3 $N\\in \\mathbb{Z}\\Rightarrow 8\\,\\,\\,\\vdots \\,\\,\\,\\left( 3x-1 \\right)\\,\\,\\Rightarrow \\left( 3x-1 \\right)\\,\\,\\in \u01af\\left( 8 \\right)$ <br\/> M\u00e0 $\u01af\\left( 8 \\right)=\\left\\{ \\pm 1;\\,\\,\\pm 2;\\,\\,\\pm 4;\\,\\,\\pm 8 \\right\\}$ <br\/> Ta c\u00f3 b\u1ea3ng sau: <br\/> <table><tr><th>$3x-1$<\/th><th>-8<\/th><th>-4<\/th><th>-2<\/th><th>-1<\/th><th>1<\/th><th>2<\/th><th>4<\/th><th>8<\/th><\/tr><tr><td>$x$<\/td><td>$\\dfrac{-7}{3}$<\/td><td>$-1$<\/td><td>$\\dfrac{-1}{3}$<\/td><td>0<\/td><td>$\\dfrac{2}{3}$<\/td><td>1<\/td><td>$\\dfrac{5}{3}$<\/td><td>3<\/td><\/tr><\/table> <br\/> Do $x\\in \\mathbb{Z}$ n\u00ean $x\\in \\{-1;0;1;3\\}$ <br\/> <span class='basic_pink'> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng l\u00e0 $-1;0;1;3$ <\/span><\/span>"}]}],"id_ques":1096},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 \u0111\u00fang nh\u1ea5t v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"],["2"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> X\u00e1c \u0111\u1ecbnh $a;b\\in \\mathbb{Z}$ bi\u1ebft r\u1eb1ng $3x=(a+b)x+2a-b$ <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $\\left\\{ \\begin{align} & a= \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & b= \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{align} \\right.$ <\/span>","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> \u0110\u1ed3ng nh\u1ea5t h\u1ec7 s\u1ed1 c\u1ee7a bi\u1ec3u th\u1ee9c v\u1ebf tr\u00e1i v\u1edbi h\u1ec7 s\u1ed1 c\u1ee7a bi\u1ec3u th\u1ee9c v\u1ebf ph\u1ea3i. <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Theo b\u00e0i: $3x=(a+b)x+2a-b$ <br\/> \u0110\u1ed3ng nh\u1ea5t h\u1ec7 s\u1ed1 bi\u1ec3u th\u1ee9c \u1edf hai v\u1ebf tr\u00ean, ta c\u00f3: <br\/> $\\left\\{ \\begin{align} & 3=a+b\\,\\,\\,\\left( 1 \\right) \\\\ & 0=2a-b\\,\\,\\,\\,\\left( 2 \\right) \\\\ \\end{align} \\right.$ <br\/> T\u1eeb (2) ta c\u00f3: $b=2a$ <br\/> Thay $b=2a$ v\u00e0o (1) ta \u0111\u01b0\u1ee3c: <br\/> $3=a+2a \\Rightarrow 3a=3 \\Rightarrow a=1$ <br\/> $\\Rightarrow b=2a=2.1=2$ <br\/> V\u1eady $a=1;b=2$ <br\/> <span class='basic_pink'> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $1;2$ <\/span><\/span>"}]}],"id_ques":1097},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00e2y \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" <span class='basic_left'> V\u1edbi $\\forall x;y \\in \\mathbb{Q}$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c sau lu\u00f4n lu\u00f4n l\u00e0 s\u1ed1 d\u01b0\u01a1ng: <br\/> $M=\\dfrac{3\\left( {{x}^{2}}+1 \\right)+{{x}^{2}}{{y}^{2}}+{{y}^{2}}-2}{{{\\left( x+y \\right)}^{2}}+5}$ <\/span>","select":["\u0110\u00fang","Sai "],"hint":"Bi\u1ebfn \u0111\u1ed5i v\u00e0 nh\u1eadn \u0111\u1ecbnh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c \u0111\u00e3 cho","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{aligned}& M=\\dfrac{3\\left( {{x}^{2}}+1 \\right)+{{x}^{2}}{{y}^{2}}+{{y}^{2}}-2}{{{\\left( x+y \\right)}^{2}}+5} \\\\ & \\,\\,\\,\\,\\,\\,=\\dfrac{3{{x}^{2}}+3+{{x}^{2}}{{y}^{2}}+{{y}^{2}}-2}{{{\\left( x+y \\right)}^{2}}+5} \\\\ & \\,\\,\\,\\,\\,=\\dfrac{3{{x}^{2}}+{{x}^{2}}{{y}^{2}}+{{y}^{2}}+1}{{{\\left( x+y \\right)}^{2}}+5} \\\\ & Do\\,\\,\\left\\{ \\begin{aligned}& {{x}^{2}}\\ge 0 \\,\\forall \\,x \\\\ & {{x}^{2}}{{y}^{2}}\\ge 0 \\, \\forall \\,x,y\\\\ & {{y}^{2}}\\ge 0 \\,\\forall \\,y \\\\ \\end{aligned} \\right.\\Rightarrow 3{{x}^{2}}+{{x}^{2}}{{y}^{2}}+{{y}^{2}}+1\\ge 1\\,\\,\\,\\,\\,\\,\\left( 1 \\right) \\\\ & Do\\,\\,{{\\left( x+y \\right)}^{2}}\\ge 0\\,\\,\\,\\Rightarrow {{\\left( x+y \\right)}^{2}}+5\\ge 5\\,\\,\\,\\,\\left( 2 \\right) \\\\ \\end{aligned}$ <br\/> T\u1eeb (1) v\u00e0 (2) suy ra $M >0$ v\u1edbi $\\forall x;y \\in \\mathbb{Q}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span><\/span>","column":2}]}],"id_ques":1098},{"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'> T\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t (GTLN) c\u1ee7a bi\u1ec3u th\u1ee9c: $B=\\dfrac{{{x}^{2}}+{{y}^{2}}+3}{{{x}^{2}}+{{y}^{2}}+2}$ <\/span>","select":["A. $Max\\,B=1\\dfrac{1}{2}$ khi $x=y=0$","B. $Max\\,B=\\dfrac{5}{4}$ khi $x=y=1$","C. $Max\\,B=\\dfrac{4}{3}$ khi $x=0;y=1$","D. $Max\\,B=\\dfrac{4}{3}$ khi $x=1;y=0$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>Ph\u00e2n t\u00edch bi\u1ec3u th\u1ee9c $B$ v\u1ec1 d\u1ea1ng $a+\\dfrac{b}{A(x)}$ v\u1edbi $a;b$ l\u00e0 h\u1eb1ng s\u1ed1, $A(x)$ l\u00e0 bi\u1ec3u th\u1ee9c ch\u1ee9a bi\u1ebfn $x$ <br\/> $B$ l\u1edbn nh\u1ea5t khi v\u00e0 ch\u1ec9 khi $A(x)$ nh\u1ecf nh\u1ea5t. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3: <br\/> $\\begin{align} & B=\\dfrac{{{x}^{2}}+{{y}^{2}}+3}{{{x}^{2}}+{{y}^{2}}+2} \\\\ & \\,\\,\\,\\,\\,\\,=\\dfrac{\\left( {{x}^{2}}+{{y}^{2}}+2 \\right)+1}{{{x}^{2}}+{{y}^{2}}+2} \\\\ & \\,\\,\\,\\,\\,\\,=1+\\dfrac{1}{{{x}^{2}}+{{y}^{2}}+2} \\\\ \\end{align}$ <br\/> $B$ l\u1edbn nh\u1ea5t khi v\u00e0 ch\u1ec9 khi $\\dfrac{1}{{{x}^{2}}+{{y}^{2}}+2}$ l\u1edbn nh\u1ea5t <br\/> $\\Leftrightarrow x^2+y^2+2$ nh\u1ecf nh\u1ea5t <br\/> Do $x^2\\ge 0;\\,y^2\\ge 0$ n\u00ean $x^2+y^2+2 \\ge 2$ <br\/> $\\Rightarrow x^2+y^2+2$ nh\u1ecf nh\u1ea5t b\u1eb1ng $2$ khi $x=y=0$ <br\/> $\\Rightarrow Max\\,B=1+\\dfrac{1}{2}=1\\dfrac{1}{2}$ khi $x=y=0$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span>","column":2}]}],"id_ques":1099},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 \u0111\u00fang nh\u1ea5t v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2"],["6"],["4"],["8"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> X\u00e1c \u0111\u1ecbnh $x;y;z;t$ l\u00e0 c\u00e1c s\u1ed1 nguy\u00ean d\u01b0\u01a1ng bi\u1ebft r\u1eb1ng $yt=48;\\,yz=24;\\,xy=12;\\,zt=32.$ <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $\\left\\{ \\begin{align} & x= \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & y= \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & z= \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & t= \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{align} \\right.$ <\/span>","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Nh\u00e2n t\u1eebng v\u1ebf c\u1ee7a $yt=48$ v\u00e0 $yz=24$ v\u1edbi nhau suy ra gi\u00e1 tr\u1ecb c\u1ee7a $y$ <br\/> T\u1eeb c\u00e1c d\u1eef ki\u1ec7n c\u1ee7a \u0111\u1ec1 b\u00e0i t\u00ecm c\u00e1c gi\u00e1 tr\u1ecb $x;z;t$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Theo \u0111\u1ec1 b\u00e0i ta c\u00f3: <br\/> $yt=48\\,\\,\\,(1)$ <br\/> $yz=24\\,\\,\\,(2)$ <br\/> $xy=12\\,\\,\\,(3)$ <br\/> $zt=32\\,\\,\\,(4)$ <br\/> Nh\u00e2n t\u1eebng v\u1ebf c\u1ee7a (1) v\u00e0 (2) v\u1edbi nhau ta \u0111\u01b0\u1ee3c: $y^2zt=48.24=1152$ <br\/> V\u00ec $zt=32$ (theo (4)) n\u00ean $y^2.32=1152 \\Rightarrow y^2=1152:32=36$ <br\/> Do $y > 0 \\Rightarrow y=6$ <br\/> Theo (1) suy ra $t=48:y=48:6=8$ (th\u1ecfa m\u00e3n) <br\/> Theo (2) suy ra $z=24:y=24:6=4$ (th\u1ecfa m\u00e3n) <br\/> Theo (3) suy ra $x=12:y=12:6=2$ (th\u1ecfa m\u00e3n) <br\/> V\u1eady $x=2;y=6;z=4;t=8$ <br\/> <span class='basic_pink'> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $2;6;4;8$ <\/span><\/span>"}]}],"id_ques":1100}],"lesson":{"save":0,"level":3}}