{"segment":[{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai7/lv3/img\/4.jpg' \/><\/center>Bi\u1ec3u th\u1ee9c $\\sqrt{\\dfrac{x-2}{x+3}}$ c\u00f3 ngh\u0129a khi: ","select":["A. $x\\ge 2$ ","B. $x\\ge 2$ ho\u1eb7c $x < -3$","C. $x < -3$","D. $-3\\le x\\le 2$"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $\\dfrac{x-2}{x+3}\\ge 0$ <br\/>Tr\u01b0\u1eddng h\u1ee3p 1: $\\left\\{ \\begin{aligned} & x-2\\ge 0 \\\\ & x+3>0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\ge 2 \\\\ & x>-3 \\\\ \\end{aligned} \\right.\\Leftrightarrow x\\ge 2$ <br\/>Tr\u01b0\u1eddng h\u1ee3p 2: $\\left\\{ \\begin{aligned} & x-2\\le 0 \\\\ & x+3<0 \\\\\\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\le 2 \\\\ & x<-3 \\\\ \\end{aligned} \\right.\\Leftrightarrow x<-3$ <br\/>V\u1eady bi\u1ec3u th\u1ee9c $\\sqrt{\\dfrac{x-2}{x+3}}$ c\u00f3 ngh\u0129a khi $x\\ge 2$ ho\u1eb7c $x < -3$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":751},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"<span class='basic_left'>V\u1edbi $x>0;\\,\\,x\\ne 1$. Cho bi\u1ec3u th\u1ee9c: <br\/>$P=\\left( \\dfrac{\\sqrt{x}}{\\sqrt{x}-1}+\\dfrac{\\sqrt{x}}{x-1} \\right)\\,$$:\\left[ \\dfrac{2}{x}-\\dfrac{2-x}{x\\left( \\sqrt{x}+1 \\right)} \\right]$ <br\/><b> C\u00e2u 1: <\/b> R\u00fat g\u1ecdn $P$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0:<\/span> ","select":["A. $-\\dfrac{x}{\\sqrt{x}+1}$ ","B. $-\\dfrac{x}{\\sqrt{x}-1}$ ","C. $\\dfrac{x}{\\sqrt{x}+1}$","D. $\\dfrac{x}{\\sqrt{x}-1}$"],"hint":"R\u00fat g\u1ecdn hai bi\u1ec3u th\u1ee9c trong ngo\u1eb7c tr\u01b0\u1edbc r\u1ed3i m\u1edbi th\u1ef1c hi\u1ec7n ph\u00e9p chia hai bi\u1ec3u th\u1ee9c","explain":"<span class='basic_left'>V\u1edbi $x>0;\\,\\,x\\ne 1$. Ta c\u00f3:<br\/>$P=\\left( \\dfrac{\\sqrt{x}}{\\sqrt{x}-1}+\\dfrac{\\sqrt{x}}{x-1} \\right)\\,$$:\\left[ \\dfrac{2}{x}-\\dfrac{2-x}{x\\left( \\sqrt{x}+1 \\right)} \\right]$<br\/>$ =\\dfrac{\\sqrt{x}\\left( \\sqrt{x}+1 \\right)+\\sqrt{x}}{\\left( \\sqrt{x}-1 \\right)\\left( \\sqrt{x}+1 \\right)}\\,$$:\\dfrac{2\\sqrt{x}+2-2+x}{x\\left( \\sqrt{x}+1 \\right)} $<br\/>$ =\\dfrac{x+2\\sqrt{x}}{\\left( \\sqrt{x}-1 \\right)\\left( \\sqrt{x}+1 \\right)}\\,$$.\\dfrac{x\\left( \\sqrt{x}+1 \\right)}{x+2\\sqrt{x}} $<br\/>$ =\\dfrac{x}{\\sqrt{x}-1}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t D<\/span><\/span>","column":2}]}],"id_ques":752},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":" <span class='basic_left'>V\u1edbi $x>0;\\,\\,x\\ne 1$ . Cho bi\u1ec3u th\u1ee9c: <br\/>$P=\\left( \\dfrac{\\sqrt{x}}{\\sqrt{x}-1}+\\dfrac{\\sqrt{x}}{x-1} \\right)\\,$$:\\left[ \\dfrac{2}{x}-\\dfrac{2-x}{x\\left( \\sqrt{x}+1 \\right)} \\right]$<br\/><b> C\u00e2u 2: <\/b> T\u00ecm $x$ khi $P=2$<\/span>","select":["A. $x=4$ ","B. $x = 2$","C. $x\\in \\varnothing$","D. $x \\in \\mathbb R$"],"hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $P=2$","explain":"<span class='basic_left'>Theo c\u00e2u 1, v\u1edbi $x>0;\\,\\,x\\ne 1$, ta c\u00f3:<br\/>$P=\\dfrac{x}{\\sqrt{x}-1}$ <br\/>Ta c\u00f3: <br\/>$\\begin{aligned} & P=2\\Leftrightarrow \\dfrac{x}{\\sqrt{x}-1}=2 \\\\ & \\Leftrightarrow \\dfrac{x}{\\sqrt{x}-1}=\\dfrac{2\\left( \\sqrt{x}-1 \\right)}{\\sqrt{x}-1} \\\\ & \\Leftrightarrow x-2\\sqrt{x}+2=0 \\\\ & \\Leftrightarrow {{\\left( \\sqrt{x}-1 \\right)}^{2}}+1=0 \\\\ \\end{aligned}$ <br\/>V\u00ec ${{\\left( \\sqrt{x}-1 \\right)}^{2}}\\ge 0 \\Rightarrow {{\\left( \\sqrt{x}-1 \\right)}^{2}}+1 > 0 \\,\\,\\,\\forall x>0;\\,\\,x\\ne 1$ <br\/>V\u1eady kh\u00f4ng c\u00f3 gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a $x$ \u0111\u1ec3 $P= 2$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t C<\/span>","column":2}]}],"id_ques":753},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["4"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":" <span class='basic_left'>V\u1edbi $x>0;\\,\\,x\\ne 1$. Cho bi\u1ec3u th\u1ee9c: <br\/>$P=\\left( \\dfrac{\\sqrt{x}}{\\sqrt{x}-1}+\\dfrac{\\sqrt{x}}{x-1} \\right)\\,$$:\\left[ \\dfrac{2}{x}-\\dfrac{2-x}{x\\left( \\sqrt{x}+1 \\right)} \\right]$<br\/><b> C\u00e2u 3: <\/b> T\u00ecm $x \\in\\,\\mathbb Z$ \u0111\u1ec3 $P\\in\\mathbb Z$ <br\/><b>\u0110\u00e1p \u00e1n:<\/b> $x=$_input_ <\/span>","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c $P$ v\u1ec1 d\u1ea1ng $m + \\dfrac{n}{f(x)}$ v\u1edbi $m, n \\in \\mathbb Z$<br\/>B\u01b0\u1edbc 2: Cho m\u1eabu c\u1ee7a ph\u00e2n th\u1ee9c \u0111\u00f3 b\u1eb1ng \u01b0\u1edbc c\u1ee7a t\u1eed <br\/>B\u01b0\u1edbc 3: T\u00ecm $x$ v\u00e0 so s\u00e1nh v\u1edbi \u0111i\u1ec1u ki\u1ec7n v\u00e0 k\u1ebft lu\u1eadn.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'>Theo c\u00e2u 1, v\u1edbi $x>0;\\,\\,x\\ne 1$, ta c\u00f3:<br\/>$P=\\dfrac{x}{\\sqrt{x}-1}=\\sqrt{x}+1+\\dfrac{1}{\\sqrt{x}-1}$<br\/>\u0110\u1ec3 $P\\in \\mathbb Z\\Leftrightarrow \\dfrac{1}{\\sqrt{x}-1}\\in \\mathbb Z\\Leftrightarrow \\sqrt{x}-1\\in \u01af(1)=\\{\\pm 1\\}$<br\/>Ta c\u00f3 b\u1ea3ng sau:<br\/><table> <tr> <th>$\\sqrt{x}-1$<\/th> <th>-1<\/th> <th>1<\/th> <\/tr> <tr> <td>$\\sqrt{x}$<\/td> <td>0<\/td> <td>2<\/td> <\/tr> <tr> <td>x<\/td> <td>lo\u1ea1i <\/td> <td>4<\/td> <\/tr><\/table><br\/>K\u1ebft h\u1ee3p v\u1edbi \u0111i\u1ec1u ki\u1ec7n $x >0$ v\u00e0 $x\\ne1$ $\\Rightarrow$ $x=4$ th\u00ec $P\\in \\mathbb Z$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $4$<\/span><\/span><\/span> "}]}],"id_ques":754},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u ","temp":"fill_the_blank_random","correct":[[["4"],["4"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":" <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c: $P=\\left( \\dfrac{\\sqrt{x}}{\\sqrt{x}-1}+\\dfrac{\\sqrt{x}}{x-1} \\right)\\,$$:\\left[ \\dfrac{2}{x}-\\dfrac{2-x}{x\\left( \\sqrt{x}+1 \\right)} \\right]$<br\/><b> C\u00e2u 4: <\/b> V\u1edbi $x > 1$, gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $P$ t\u00ecm \u0111\u01b0\u1ee3c l\u00e0 $P=$_input_khi $x=$ _input_<\/span>","hint":"\u0110\u01b0a $P$ v\u1ec1 d\u1ea1ng $f(x)+\\dfrac{a}{f(x)}+b$ trong \u0111\u00f3 $a>0$, $f(x) >0$. Sau \u0111\u00f3, \u00e1p d\u1ee5ng B\u0110T Cauchy","explain":"<span class='basic_left'>Theo c\u00e2u 1, v\u1edbi $x>0;\\,\\,x\\ne 1$, ta c\u00f3:<br\/> $P=\\dfrac{x}{\\sqrt{x}-1}=\\sqrt{x}+1+\\dfrac{1}{\\sqrt{x}-1}\\,$$=\\sqrt{x}-1+\\dfrac{1}{\\sqrt{x}-1}+2$<br\/>V\u00ec $x > 1 \\Rightarrow \\sqrt{x}-1>0$ <br\/>\u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c Cauchy cho hai s\u1ed1 kh\u00f4ng \u00e2m $\\sqrt{x}-1$ v\u00e0 $\\dfrac{1}{\\sqrt{x}-1}$, ta c\u00f3: <br\/>$\\sqrt{x}-1+\\dfrac{1}{\\sqrt{x}-1}\\,$$\\ge 2\\sqrt{\\left( \\sqrt{x}-1 \\right).\\dfrac{1}{\\sqrt{x}-1}}=2$ <br\/>Suy ra, $ P\\ge 2+2\\ge 4$. Do \u0111\u00f3, <br\/>$\\begin{align} {{P}_{\\min }}=4& \\Leftrightarrow \\sqrt{x}-1=\\dfrac{1}{\\sqrt{x}-1} \\\\ & \\Leftrightarrow {{\\left( \\sqrt{x}-1 \\right)}^{2}}=1 \\\\ & \\Leftrightarrow \\sqrt{x}-1=1 \\,\\, (\\text {V\u00ec} x > 1)\\\\ & \\Leftrightarrow \\sqrt{x}=2 \\\\ & \\Leftrightarrow x=4 \\,\\, (\\text {(th\u1ecfa m\u00e3n)})\\\\ \\end{align}$ <br\/>V\u1eady gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $P$ l\u00e0 $4$ khi $x=4$<br\/> <span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $4; 4$ <\/span><\/span><\/span>"}]}],"id_ques":755},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u $> , < , =$ th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank_random","correct":[[[">"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai7/lv3/img\/5.jpg' \/><\/center> Cho hai bi\u1ec3u th\u1ee9c $A=1+\\sqrt{12}+\\sqrt{37}$ v\u00e0 $B=6\\sqrt{3}$ <br\/>So s\u00e1nh: $A$ _input_ $B$","hint":"So s\u00e1nh ${{\\left( 1+\\sqrt{37} \\right)}^{2}}$ v\u00e0 $(4\\sqrt 3)^2$","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$A=1+\\sqrt{12}+\\sqrt{37}\\,$$=1+2\\sqrt{3}+\\sqrt{37}$<br\/>$B=4\\sqrt 3 +2\\sqrt 3$<br\/> ${{\\left( 1+\\sqrt{37} \\right)}^{2}}=38+2\\sqrt{37}$ <br\/>${{\\left( 4\\sqrt{3} \\right)}^{2}}=48=38+2\\sqrt{25}$ <br\/>Suy ra ${{\\left( 4\\sqrt{3} \\right)}^{2}}<{{\\left( 1+\\sqrt{37} \\right)}^{2}}\\,$$\\Leftrightarrow 4\\sqrt{3}<1+\\sqrt{37}\\,$$\\Leftrightarrow 6\\sqrt{3}<1+2\\sqrt{3}+\\sqrt{37}$ <br\/>Do \u0111\u00f3 $A > B$ <br\/> <span class='basic_pink'>V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0 $>$ <\/span><\/span><\/span>"}]}],"id_ques":756},{"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":" T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{\\sqrt{{{x}^{2}}+2x-3}}{\\sqrt{x-1}}=3+x$ l\u00e0: ","select":["A. $S=\\{\\varnothing \\}$ ","B. $S=\\{-2;-3\\}$","C. $S=\\mathbb R$"],"hint":"Ph\u00e2n t\u00edch bi\u1ec3u th\u1ee9c trong c\u0103n c\u1ee7a t\u1eed th\u1ee9c b\u00ean v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed r\u1ed3i r\u00fat g\u1ecdn v\u00e0 \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng $\\sqrt {f(x)}= g(x)$.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: T\u00ecm \u0111i\u1ec1u ki\u1ec7n c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh <br\/>B\u01b0\u1edbc 2: Ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed<br\/>B\u01b0\u1edbc 3:Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh v\u00e0 t\u00ecm nghi\u1ec7m<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<br\/><\/span>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh <br\/>$\\left\\{ \\begin{aligned} & x-1>0 \\\\ & {{x}^{2}}+2x-3\\ge 0 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow \\left\\{ \\begin{aligned} & x>1 \\\\ & \\left( x-1 \\right)\\left( x+3 \\right)\\ge 0 \\\\ \\end{aligned} \\right.\\,$$\\Leftrightarrow x>1$ . <br\/>$\\begin{aligned} & \\,\\,\\,\\,\\dfrac{\\,\\sqrt{{{x}^{2}}+2x-3}}{\\sqrt{x-1}}=3+x \\\\ & \\Leftrightarrow \\dfrac{\\sqrt{\\left( x-1 \\right)\\left( x+3 \\right)}}{\\sqrt{x-1}}=x+3 \\\\ & \\Leftrightarrow \\sqrt{x+3}=x+3 \\\\ & \\Leftrightarrow x+3\\,\\,\\,\\,\\,={{\\left( x+3 \\right)}^{2}} \\\\ & \\Leftrightarrow \\left( x+3 \\right)\\left( x+2 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=-3\\,\\,\\,\\,\\left( lo\u1ea1i \\right) \\\\ & x=-2\\,\\,\\,\\,\\left( lo\u1ea1i \\right) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t A<\/span><\/span>","column":3}]}],"id_ques":757},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai7/lv3/img\/9.jpg' \/><\/center>V\u1edbi $x > y$ v\u00e0 $x.y = 1$ ta c\u00f3: $\\dfrac{{{x}^{2}}+{{y}^{2}}}{x-y}\\ge 2\\sqrt{2}$ ","select":["A. \u0110\u00fang ","B. Sai"],"hint":"Ch\u1ee9ng minh b\u1ea5t \u0111\u1eb3ng th\u1ee9c ban \u0111\u1ea7u \u0111\u00fang. Ta quy \u0111\u1ed3ng hai v\u1ebf r\u1ed3i kh\u1eed m\u1eabu, sau \u0111\u00f3 th\u00eam b\u1edbt $2=2xy$ v\u00e0o \u0111\u1ec3 xu\u1ea5t hi\u1ec7n h\u1eb1ng \u0111\u1eb3ng th\u1ee9c.","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\dfrac{{{x}^{2}}+{{y}^{2}}}{x-y}\\ge 2\\sqrt{2}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 1 \\right)$<br\/>$\\Leftrightarrow {{x}^{2}}+{{y}^{2}}\\,$$-2\\sqrt{2}\\left( x-y \\right)\\ge 0$ (do $x > y$)<br\/>$ \\Leftrightarrow {{x}^{2}}+{{y}^{2}}-2+2-2\\sqrt{2}\\left( x-y \\right)\\ge 0 $<br\/>$ \\Leftrightarrow {{x}^{2}}+{{y}^{2}}-2xy-2\\sqrt{2}\\left( x-y \\right)+\\,$$2\\ge 0\\,\\,\\,\\,\\,(V\u00ec\\,\\,x.y=1) $<br\/>$ \\Leftrightarrow {{\\left( x-y \\right)}^{2}}-2\\sqrt{2}\\left( x-y \\right)\\,$$+\\sqrt{{{2}^{2}}}\\ge 0 $<br\/>$ \\Leftrightarrow {{\\left( x-y-\\sqrt{2} \\right)}^{2}}\\ge 0\\,\\,\\,\\,\\,\\left( * \\right)$<br\/>B\u1ea5t \u0111\u1eb3ng th\u1ee9c (*) lu\u00f4n \u0111\u00fang $\\Rightarrow $ B\u1ea5t \u0111\u1eb3ng th\u1ee9c (1) lu\u00f4n \u0111\u00fang<br\/>Do \u0111\u00f3 kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0110\u00fang <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t l\u00e0 A<\/span><br\/><i>Nh\u1eadn x\u00e9t:<\/i> Ph\u01b0\u01a1ng ph\u00e1p ch\u1ee9ng minh b\u1ea5t \u0111\u1eb3ng th\u1ee9c: S\u1eed d\u1ee5ng ph\u00e9p bi\u1ebfn \u0111\u1ed5i t\u01b0\u01a1ng \u0111\u01b0\u01a1ng d\u1eabn \u0111\u1ebfn m\u1ed9t \u0111i\u1ec1u lu\u00f4n \u0111\u00fang th\u00ec b\u1ea5t \u0111\u1eb3ng th\u1ee9c ban \u0111\u1ea7u c\u0169ng lu\u00f4n \u0111\u00fang.<\/span><\/span>","column":2}]}],"id_ques":758},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\sqrt{2}$","B. $2\\sqrt{2}$","C. $3\\sqrt{2}$"],"ques":"V\u1edbi $\\dfrac{1}{6}\\le x\\le \\dfrac{1}{3}$ . R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c sau<br\/>$ A=\\sqrt{3x+\\sqrt{6x-1}}\\,$$+\\sqrt{3x-\\sqrt{6x-1}}$<br\/>\u0110\u00e1p \u00e1n: $A=$?","hint":"Nh\u00e2n $\\sqrt{2}$ v\u00e0o hai v\u1ebf r\u1ed3i \u0111\u01b0a $\\sqrt{2}$ v\u00e0o trong d\u1ea5u c\u0103n \u0111\u1ec3 chuy\u1ec3n bi\u1ec3u th\u1ee9c d\u01b0\u1edbi c\u0103n v\u1ec1 d\u1ea1ng $(A \\pm B)^2$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: Nh\u00e2n $\\sqrt{2}$ v\u00e0o hai v\u1ebf r\u1ed3i \u0111\u01b0a $\\sqrt{2}$ v\u00e0o trong d\u1ea5u c\u0103n \u0111\u1ec3 chuy\u1ec3n bi\u1ec3u th\u1ee9c d\u01b0\u1edbi c\u0103n v\u1ec1 d\u1ea1ng $(A \\pm B)^2$<br\/>B\u01b0\u1edbc 2: \u0110\u01b0a bi\u1ec3u th\u1ee9c trong c\u0103n v\u1ec1 d\u1ea1ng $(a \\pm b)^2$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'>$ A=\\sqrt{3x+\\sqrt{6x-1}}\\,$$+\\sqrt{3x-\\sqrt{6x-1}}$<br\/>$ A\\sqrt{2}=\\sqrt{6x+2\\sqrt{6x-1}}\\,$$+\\sqrt{6x-2\\sqrt{6x-1}} $<br\/>$ A\\sqrt{2}=\\sqrt{6x-1+2\\sqrt{6x-1}+1}\\,$$+\\sqrt{6x-1-2\\sqrt{6x-1}+1} $<br\/>$A\\sqrt{2}=\\sqrt{{{\\left( \\sqrt{6x-1}+1 \\right)}^{2}}}\\,$$+\\sqrt{{{\\left( \\sqrt{6x-1}-1 \\right)}^{2}}}$<br\/>$ A\\sqrt{2}=\\left| \\sqrt{6x-1}+1 \\right|+\\left| \\sqrt{6x-1}-1 \\right|$<br\/>V\u00ec $\\dfrac{1}{6}\\le x\\le \\dfrac{1}{3}$ n\u00ean $\\sqrt{6x-1}-1<0\\,$$\\Rightarrow \\left| \\sqrt{6x-1}-1 \\right|=1-\\sqrt{6x-1}$ <br\/> $A\\sqrt{2}\\,$$=\\sqrt{6x-1}+1+1-\\sqrt{6x-1}=2\\,$$\\Rightarrow A=2:\\sqrt 2 =\\sqrt{2}$ <\/span> <\/span><\/span>"}]}],"id_ques":759},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank_random","correct":[[["0"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":" V\u1edbi $a > 0,\\, b > 0$. R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c<br\/>$\\dfrac{{{\\left( \\sqrt{a}-\\sqrt{b} \\right)}^{2}}+4\\sqrt{ab}}{\\sqrt{a}+\\sqrt{b}}\\,$$-\\dfrac{a\\sqrt{b}-b\\sqrt{a}}{\\sqrt{ab}}-2\\sqrt{b}=$ _input_","hint":"Th\u1ef1c hi\u1ec7n c\u00e1c ph\u00e9p bi\u1ebfn \u0111\u1ed5i \u0111\u01a1n gi\u1ea3n v\u1ec1 h\u1eb1ng \u0111\u1eb3ng th\u1ee9c, th\u1eeba s\u1ed1 chung,... \u0111\u1ec3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$ \\,\\,\\,\\,\\dfrac{{{\\left( \\sqrt{a}-\\sqrt{b} \\right)}^{2}}+4\\sqrt{ab}}{\\sqrt{a}+\\sqrt{b}}\\,$$-\\dfrac{a\\sqrt{b}-b\\sqrt{a}}{\\sqrt{ab}}-2\\sqrt{b} $<br\/>$ =\\dfrac{a-2\\sqrt{ab}+b+4\\sqrt{ab}}{\\sqrt{a}+\\sqrt{b}}\\,$$-\\dfrac{\\sqrt{ab}\\left( \\sqrt{a}-\\sqrt{b} \\right)}{\\sqrt{ab}}-2\\sqrt{b} $<br\/>$ \\begin{align}&=\\dfrac{{{\\left( \\sqrt{a}+\\sqrt{b} \\right)}^{2}}}{\\sqrt{a}+\\sqrt{b}}-\\sqrt{a}+\\sqrt{b}-2\\sqrt{b} \\\\ & =\\sqrt{a}+\\sqrt{b}-\\sqrt{a}-\\sqrt{b} \\\\ & =0 \\\\ \\end{align}$ <br\/> <span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $0$<\/span><\/span><\/span>"}]}],"id_ques":760}],"lesson":{"save":0,"level":3}}