{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $10\\sqrt{3}$","B. $8\\sqrt{3}$","C. $5\\sqrt{3}$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai5/lv1/img\/2.jpg' \/><\/center>Th\u1ef1c hi\u1ec7n ph\u00e9p t\u00ednh:<br\/> $5\\sqrt{12}-4\\sqrt{3}+\\sqrt{48} =$?","hint":"\u00c1p d\u1ee5ng: Quy t\u1eafc \u0111\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> \u00c1p d\u1ee5ng: Quy t\u1eafc \u0111\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n <br\/><b>B\u01b0\u1edbc 2:<\/b> C\u1ed9ng tr\u1eeb c\u00e1c c\u0103n th\u1ee9c \u0111\u1ed3ng d\u1ea1ng <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/>$\\begin{align} & \\,\\,\\,5\\sqrt{12}-4\\sqrt{3}+\\sqrt{48} \\\\ & \\,\\,\\,=5\\sqrt{4.3}-4\\sqrt{3}+\\sqrt{16.3} \\\\ & =10\\sqrt{3}-4\\sqrt{3}+4\\sqrt{3} \\\\ & =10\\sqrt{3} \\\\ \\end{align}$<\/span> <\/span>"}]}],"id_ques":611},{"time":24,"part":[{"title":"\u0110i\u1ec1n bi\u1ec3u th\u1ee9c th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["1"],["a"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai5/lv1/img\/9.jpg' \/><\/center> $\\left( \\sqrt{a}-\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\right)$$\\left( \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} +\\sqrt{a}+1 \\right)\\,$$=a\\sqrt{a}-1$ ","hint":"Bi\u1ebfn \u0111\u1ed5i v\u1ebf ph\u1ea3i: \u00c1p d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $ a^3-b^3$ ","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$a\\sqrt{a}-1=(\\sqrt{a})^3-1^3=(\\sqrt{a}-1)(a+\\sqrt{a}+1)$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $1$ v\u00e0 $a$ <\/span> <br\/> <b> Ghi nh\u1edb: <\/b> V\u1edbi $a\\ge 0$, $a\\sqrt a = (\\sqrt a)^3$ <\/span>"}]}],"id_ques":612},{"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":[[1]],"list":[{"point":5,"ques":"<span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $B=\\dfrac{\\sqrt{x}}{\\sqrt{x}-3}+\\dfrac{2\\sqrt{x}-24}{x-9}$<br\/><b> C\u00e2u 1: <\/b> R\u00fat g\u1ecdn $B$<\/span>","select":["A. $B=\\dfrac{\\sqrt{x}+8}{\\sqrt{x}+3}$","B. $B=\\dfrac{\\sqrt{x}+8}{\\sqrt{x}-3}$","C. $B=\\dfrac{\\sqrt{x}-8}{\\sqrt{x}+3}$ ","D. $B=\\dfrac{\\sqrt{x}-8}{\\sqrt{x}-3}$"],"hint":"Ph\u00e2n t\u00edch c\u00e1c m\u1eabu th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed r\u1ed3i quy \u0111\u1ed3ng v\u00e0 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 bi\u1ec3u th\u1ee9c c\u00f3 ngh\u0129a <br\/>B\u01b0\u1edbc 2: T\u00ecm m\u1eabu th\u1ee9c chung v\u00e0 quy \u0111\u1ed3ng<br\/>B\u01b0\u1edbc 3: Th\u1ef1c hi\u1ec7n c\u00e1c ph\u00e9p bi\u1ebfn \u0111\u1ed5i kh\u00e1c \u0111\u1ec3 r\u00fat g\u1ecdn $B$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n: $x\\ge 0;\\,\\,x\\ne 9$<br\/>Ta c\u00f3:<br\/>$\\begin{aligned} B&=\\dfrac{\\sqrt{x}}{\\sqrt{x}-3}+\\dfrac{2\\sqrt{x}-24}{x-9} \\\\ & =\\dfrac{\\sqrt{x}\\left( \\sqrt{x}+3 \\right)+2\\sqrt{x}-24}{\\left( \\sqrt{x}-3 \\right)\\left( \\sqrt{x}+3 \\right)} \\\\ & =\\dfrac{x+3\\sqrt{x}+2\\sqrt{x}-24}{\\left( \\sqrt{x}-3 \\right)\\left( \\sqrt{x}+3 \\right)} \\\\& = \\dfrac{x + 5\\sqrt{x} - 24}{\\left( \\sqrt{x}-3 \\right)\\left( \\sqrt{x}+3 \\right)}\\\\&=\\dfrac{x - 3\\sqrt{x}+ 8\\sqrt{x} - 24}{\\left( \\sqrt{x}-3 \\right)\\left( \\sqrt{x}+3 \\right)} \\\\&= \\dfrac{\\sqrt{x}(\\sqrt{x} - 3) + 8(\\sqrt{x} - 3) }{\\left( \\sqrt{x}-3 \\right)\\left( \\sqrt{x}+3 \\right)} \\\\ & =\\dfrac{\\left( \\sqrt{x}+8 \\right)\\left( \\sqrt{x}-3 \\right)}{\\left( \\sqrt{x}-3 \\right)\\left( \\sqrt{x}+3 \\right)} \\\\ & =\\dfrac{\\sqrt{x}+8}{\\sqrt{x}+3} \\\\ \\end{aligned}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A <\/span> <\/span>","column":2}]}],"id_ques":613},{"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":5,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $B=\\dfrac{\\sqrt{x}}{\\sqrt{x}-3}+\\dfrac{2\\sqrt{x}-24}{x-9}$<br\/><b> C\u00e2u 2: <\/b>V\u1edbi $B= 2$ th\u00ec $x=$_input_ <\/span>","hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $B=2$ \u0111\u1ec3 t\u00ecm $x$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: Bi\u1ebfn \u0111\u1ed5i ph\u01b0\u01a1ng tr\u00ecnh $B=2$ v\u1ec1 d\u1ea1ng $\\sqrt A=B\\Leftrightarrow A=B^2\\,(B\\ge0)$. <br\/>B\u01b0\u1edbc 2: Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh<br\/>B\u01b0\u1edbc 3: K\u1ebft h\u1ee3p v\u1edbi \u0111i\u1ec1u ki\u1ec7n c\u1ee7a b\u00e0i to\u00e1n \u0111\u1ec3 k\u1ebft lu\u1eadn nghi\u1ec7m <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Theo c\u00e2u 1, v\u1edbi $x\\ge 0;\\,\\,x\\ne 9$, ta c\u00f3: $B=\\dfrac{\\sqrt{x}+8}{\\sqrt{x}+3}$ <br\/> Thay $B=2$ ta c\u00f3:<br\/>$\\begin{aligned} & \\dfrac{\\sqrt{x}+8}{\\sqrt{x}+3}=2 \\\\ \\Rightarrow &\\sqrt{x}+8=2\\sqrt{x}+6 \\\\ \\Leftrightarrow &\\sqrt{x}=2 \\\\ \\Leftrightarrow &x=4\\,\\,\\text {(th\u1ecfa m\u00e3n)} \\\\ \\end{aligned}$ <br\/>V\u1eady \u0111\u1ec3 $B= 2$ th\u00ec $x= 4$<br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $4$ <\/span><\/span>"}]}],"id_ques":614},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{12}{7}$","B. $\\dfrac{6}{7}$","C. $\\dfrac{5}{7}$"],"ques":"<span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $B=\\dfrac{\\sqrt{x}}{\\sqrt{x}-3}+\\dfrac{2\\sqrt{x}-24}{x-9}$<br\/>V\u1edbi $x=16$ th\u00ec $B=$?<\/span>","hint":"Thay $x= 16$ v\u00e0o bi\u1ec3u th\u1ee9c $B$ \u0111\u00e3 \u0111\u01b0\u1ee3c r\u00fat g\u1ecdn.","explain":"<span class='basic_left'>Theo c\u00e2u 1, v\u1edbi $x\\ge 0;\\,\\,x\\ne 9$, ta c\u00f3: $B=\\dfrac{\\sqrt{x}+8}{\\sqrt{x}+3}$ <br\/>Thay $x= 16$ v\u00e0o bi\u1ec3u th\u1ee9c $B$ \u0111\u00e3 r\u00fat g\u1ecdn, ta c\u00f3:<br\/> $B=\\dfrac{\\sqrt{16}+8}{\\sqrt{16}+3}=\\dfrac{12}{7}$ <br\/>V\u1eady $x = 16$ th\u00ec $B = \\dfrac{12}{7}$"}]}],"id_ques":615},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u $>, <, =$ 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":[[[">"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai5/lv1/img\/5.jpg' \/><\/center><span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $B=\\dfrac{\\sqrt{x}}{\\sqrt{x}-3}+\\dfrac{2\\sqrt{x}-24}{x-9}$<br\/><b> C\u00e2u 4: <\/b> So s\u00e1nh $B$ v\u1edbi $1$<br\/><b>\u0110\u00e1p \u00e1n:<\/b> $B$ _input_ $1$<\/span> ","hint":"X\u00e9t hi\u1ec7u $B - 1$ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: X\u00e9t hi\u1ec7u $B-1$<br\/>B\u01b0\u1edbc 2: Ch\u1ee9ng minh $B - 1 > 0$<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'>Theo c\u00e2u 1, v\u1edbi $x\\ge 0;\\,\\,x\\ne 9$, ta c\u00f3: $B=\\dfrac{\\sqrt{x}+8}{\\sqrt{x}+3}$ <br\/>X\u00e9t hi\u1ec7u:<br\/> $B -1=\\dfrac{\\sqrt{x}+8}{\\sqrt{x}+3}-1=\\dfrac{5}{\\sqrt{x}+3}$ <br\/>Nh\u1eadn x\u00e9t: <br\/>$\\left. \\begin{aligned} & 5>0 \\\\ & \\sqrt{x}+3>0\\,\\,\\forall x\\ge 0 \\\\ \\end{aligned} \\right\\}\\,$$\\Rightarrow \\dfrac{5}{\\sqrt{x}+3}>0\\Leftrightarrow B-1>0\\Leftrightarrow B>1$ <br\/><span class='basic_pink'> V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0 $>$ <\/span><\/span><\/span>"}]}],"id_ques":616},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $-\\sqrt{5x}$","B. $-2\\sqrt{5x}$","C. $-3\\sqrt{5x}$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai5/lv1/img\/2.jpg' \/><\/center>V\u1edbi $x \\ge 0$. R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c<br\/> $\\sqrt{5x}-7\\sqrt{5x}+\\sqrt{125x}=$?","hint":"Bi\u1ebfn \u0111\u1ed5i v\u1ec1 c\u00e1c c\u0103n th\u1ee9c \u0111\u1ed3ng d\u1ea1ng","explain":"<span class='basic_left'>V\u1edbi $x\\ge 0$ ta c\u00f3:<br\/>$\\begin{align}&\\,\\,\\,\\,\\sqrt{5x}-7\\sqrt{5x}+\\sqrt{125x}\\\\&=\\sqrt{5x}-7\\sqrt{5x}+5\\sqrt{5x}\\\\&=(1-7+5)\\sqrt{5x}\\\\&=-\\sqrt{5x}\\\\ \\end{align}$<\/span>"}]}],"id_ques":617},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $2-\\sqrt{3}$","B. $2$","C. $2+\\sqrt{3}$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai5/lv1/img\/4.jpg' \/><\/center>Th\u1ef1c hi\u1ec7n ph\u00e9p t\u00ednh <br\/> $\\sqrt{7-4\\sqrt{3}}=$?","hint":"Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c trong c\u0103n v\u1ec1 $(A-B)^2$ ","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{align}&\\sqrt{7-4\\sqrt{3}}\\\\&=\\sqrt{2^2-2.2.\\sqrt{3}+(\\sqrt{3})^2}\\\\&=\\sqrt{(2-\\sqrt{3})^2}\\\\&=|2-\\sqrt{3}|\\\\&=2-\\sqrt{3}\\,\\,\\,\\,(\\text {V\u00ec}\\,\\,2\\,>\\sqrt{3})\\\\ \\end{align}$ <br\/><\/span>"}]}],"id_ques":618},{"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":[[["2","-3"],["-3","2"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $\\sqrt{4{{x}^{2}}+4x+1}=5$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${_input_;_input_}","hint":"Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c trong c\u0103n v\u1ec1 d\u1ea1ng $(A+B)^2$ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span><b>B\u01b0\u1edbc 1:<\/b> Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c trong c\u0103n v\u1ec1 d\u1ea1ng $(A+B)^2$<br\/><b>B\u01b0\u1edbc 2:<\/b> Khai c\u0103n<br\/><b>B\u01b0\u1edbc 3:<\/b> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/>$\\begin{aligned} & \\sqrt{4{{x}^{2}}+4x+1}=5 \\\\ & \\Leftrightarrow \\sqrt{{{\\left( 2x+1 \\right)}^{2}}}=5 \\\\ & \\Leftrightarrow \\left| 2x+1 \\right|\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=5 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & 2x+1=5 \\\\ & 2x+1=-5 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & x=2 \\\\ & x=-3 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{2; -3\\}$ <br\/><span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2; -3$ <\/span> <\/span> "}]}],"id_ques":619},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\sqrt{5}$","B. $2\\sqrt{5}$","C. $3\\sqrt{5}$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai5/lv1/img\/2.jpg' \/><\/center> $(x - 2)^2=9-4\\sqrt{5}.$ <\/br>$ x= ?$","hint":"Bi\u1ebfn \u0111\u1ed5i v\u1ebf ph\u1ea3i: \u00c1p d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $(a-b)^2$ ","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$9-4\\sqrt{5}=(\\sqrt{5})^2-2.2.\\sqrt{5}+2^2\\,$$=(\\sqrt{5}-2)^2$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 $\\sqrt{5}$<\/span><br\/><i>L\u01b0u \u00fd:<\/i> \u0110\u1ec3 ph\u00e2n t\u00edch bi\u1ec3u th\u1ee9c $a\\pm 2b\\sqrt c $ v\u1edbi $b^2 + c =a$ v\u1ec1 d\u1ea1ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $(A\\pm B)^2$ ta th\u01b0\u1eddng x\u00e1c \u0111\u1ecbnh \u0111\u01b0\u1ee3c: $A=b$ v\u00e0 $B=\\sqrt c$. Khi \u0111\u00f3 ta c\u00f3: $a\\pm 2b\\sqrt c =(b\\pm \\sqrt c)^2$<\/span>"}]}],"id_ques":620},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"select":["A. $\\sqrt{3}+ 2$","B. $\\sqrt{3}-2$","C. $\\sqrt{3}$"],"ques":"T\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c $B=\\sqrt{4a^2-4a+1}-\\sqrt{a^2+2a+1}$ t\u1ea1i $ a=\\sqrt{3}$ <br\/>\u0110\u00e1p s\u1ed1: $B=$?","hint":"R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $B$ r\u1ed3i thay $a=\\sqrt 3$ v\u00e0o bi\u1ec3u th\u1ee9c $B$ \u0111\u00e3 r\u00fat g\u1ecdn. ","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 trong c\u0103n v\u1ec1 $(A \\pm B)^2$<br\/>B\u01b0\u1edbc 2: Khai c\u0103n v\u00e0 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $B$ <br\/>B\u01b0\u1edbc 3: Thay $a =\\sqrt{3}$ v\u00e0o bi\u1ec3u th\u1ee9c $B$ \u0111\u1ec3 t\u00ednh gi\u00e1 tr\u1ecb<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/> $\\begin{align} B&=\\sqrt{4a^2-4a+1}-\\sqrt{a^2+2a+1}\\\\&=\\sqrt{(2a-1)^2}-\\sqrt{(a+1)^2}\\\\&=|2a-1|-|a+1|\\end{align} $<br\/> Thay $a=\\sqrt{3}$ bi\u1ec3u th\u1ee9c $B$ \u0111\u00e3 r\u00fat g\u1ecdn, ta c\u00f3:<br\/>$ B= |2\\sqrt{3}-1|-|\\sqrt{3}+1|\\,$<br\/>$=2\\sqrt{3}-1-\\sqrt{3}-1\\,$<br\/>$=\\sqrt{3}-2$<br\/><\/span> "}]}],"id_ques":621},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{2}{x-1}$","B. $\\dfrac{2}{x+1}$","C. $\\dfrac{2}{x+2}$"],"ques":"<span class='basic_left'>Cho bi\u1ec3u th\u1ee9c: $A=\\dfrac{1}{\\sqrt{x}-1}-\\dfrac{1}{\\sqrt{x}+1}$<br\/> <b>C\u00e2u 1: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $A$<br\/>\u0110\u00e1p \u00e1n: $A=$?<\/span>","hint":"Quy \u0111\u1ed3ng r\u1ed3i r\u00fat g\u1ecdn $A$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 bi\u1ec3u th\u1ee9c c\u00f3 ngh\u0129a <br\/>B\u01b0\u1edbc 2: T\u00ecm m\u1eabu th\u1ee9c chung v\u00e0 quy \u0111\u1ed3ng c\u00e1c ph\u00e2n th\u1ee9c<br\/>B\u01b0\u1edbc 3: R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $A$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ge 0;\\,\\,x\\ne 1$ <br\/>Ta c\u00f3:<br\/>$\\begin{aligned} A& =\\dfrac{1}{\\sqrt{x}-1}-\\dfrac{1}{\\sqrt{x}+1} \\\\ & =\\dfrac{\\sqrt{x}+1-\\left( \\sqrt{x}-1 \\right)}{x-1} \\\\ & =\\dfrac{\\sqrt{x}+1-\\sqrt{x}+1}{x-1}=\\dfrac{2}{x-1} \\\\ \\end{aligned}$ <br\/><\/span><\/span>"}]}],"id_ques":622},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{\\sqrt{3}}{3}$","B. $\\dfrac{\\sqrt{3}}{4}$","C. $\\dfrac{\\sqrt{3}}{5}$"],"ques":"<span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $\\,A=\\dfrac{1}{\\sqrt{x}-1}-\\dfrac{1}{\\sqrt{x}+1}$<br\/><b> C\u00e2u 2: <\/b> V\u1edbi $x= 2\\sqrt{3}+1$ th\u00ec $A=$?<\/span> ","hint":"Thay $x=2\\sqrt{3}+1$ v\u00e0o bi\u1ec3u th\u1ee9c $A$ \u0111\u00e3 r\u00fat g\u1ecdn.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: Thay $x=2\\sqrt{3}+1$ v\u00e0o bi\u1ec3u th\u1ee9c $A$ \u0111\u00e3 r\u00fat g\u1ecdn. <br\/>B\u01b0\u1edbc 2: \u00c1p d\u1ee5ng: Quy t\u1eafc tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu:<br\/>V\u1edbi c\u00e1c bi\u1ec3u th\u1ee9c $A, B$ m\u00e0 $B > 0$ ta c\u00f3 $\\dfrac{A}{\\sqrt{B}}=\\dfrac{A\\sqrt{B}}{B}$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>V\u1edbi $x\\ge 0;\\,\\,x\\ne 1$, ta c\u00f3:<br\/> $A=\\dfrac{2}{x-1}$ <br\/>Thay $x= 2\\sqrt{3}+1$ v\u00e0o bi\u1ec3u th\u1ee9c $A$, ta c\u00f3: <br\/>$A=\\dfrac{2}{x-1}=\\dfrac{2}{2\\sqrt{3}+1-1}=\\dfrac{1}{\\sqrt{3}}\\,$$=\\dfrac{\\sqrt{3}}{3}$ <br\/><\/span><\/span>"}]}],"id_ques":623},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["2"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $\\,A=\\dfrac{1}{\\sqrt{x}-1}-\\dfrac{1}{\\sqrt{x}+1}$<br\/><b> C\u00e2u 3: <\/b>V\u1edbi $A= x$ th\u00ec $x=$_input_ <\/span>","hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $A=x$, s\u1eed d\u1ee5ng k\u1ebft qu\u1ea3 c\u00e2u 1.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: Bi\u1ebfn \u0111\u1ed5i ph\u01b0\u01a1ng tr\u00ecnh $A=x$ v\u1ec1 d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh t\u00edch<br\/>B\u01b0\u1edbc 2: Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh. K\u1ebft h\u1ee3p v\u1edbi \u0111i\u1ec1u ki\u1ec7n c\u1ee7a b\u00e0i to\u00e1n \u0111\u1ec3 t\u00ecm nghi\u1ec7m <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Theo c\u00e2u 1, v\u1edbi $x\\ge 0;\\,\\,x\\ne 1$, ta c\u00f3:<br\/> $A=\\dfrac{2}{x-1}$ <br\/>Thay $A=x$. Ta c\u00f3:<br\/>$ \\,\\,\\,\\,\\dfrac{2}{x-1}=x\\,$<br\/> $\\begin{aligned} & \\Rightarrow {{x}^{2}}-x\\,\\,\\,\\,\\,\\,\\,=2 \\\\ & \\Leftrightarrow {{x}^{2}}-x-2=0 \\\\&\\Leftrightarrow x^2 + x - 2x - 2 = 0\\\\&\\Leftrightarrow x(x + 1) - 2(x + 1) = 0 \\\\ & \\Leftrightarrow \\left( x+1 \\right)\\left( x-2 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x+1=0 \\\\ & x-2=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & x=-1\\,\\,\\,\\text{(lo\u1ea1i)} \\\\ & x=2\\,\\,\\,\\,\\,\\,\\,\\text {(th\u1ecfa m\u00e3n)} \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{2\\}$<br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2$ <\/span><\/span>"}]}],"id_ques":624},{"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":[[["2"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $A=\\dfrac{1}{\\sqrt{x}-1}-\\dfrac{1}{\\sqrt{x}+1}$<br\/><b> C\u00e2u 4:<\/b> Gi\u00e1 tr\u1ecb nguy\u00ean nh\u1ecf nh\u1ea5t c\u1ee7a $x$ \u0111\u1ec3 $A > 0$ l\u00e0 $x=$_input_ <\/span>","hint":"Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh $A>0$ ","explain":"<span class='basic_left'>V\u1edbi $x\\ge 0;\\,\\,x\\ne 1$, ta c\u00f3:<br\/> $A=\\dfrac{2}{x-1}$ <br\/>V\u1edbi $A>0$ ta c\u00f3:<br\/>$ \\dfrac{2}{x-1}>0$<br\/>V\u00ec $2>0$n\u00ean $\\dfrac{2}{x-1}>0\\Leftrightarrow x-1>0\\Leftrightarrow x>1$ <br\/>M\u00e0 $x\\ge 0;\\,\\,x\\ne 1$ <br\/>V\u1eady gi\u00e1 tr\u1ecb nguy\u00ean nh\u1ecf nh\u1ea5t c\u1ee7a $x $ \u0111\u1ec3 $A > 0$ l\u00e0 $x = 2$<br\/><span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2$ <\/span>"}]}],"id_ques":625},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\sqrt{a}$","B. $a$","C. $2\\sqrt{a}$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai5/lv1/img\/2.jpg' \/><\/center> $(\\sqrt{a}-1)(x+1)=a-1$. <\/br> $x$ =? ","hint":"Bi\u1ebfn \u0111\u1ed5i v\u1ebf ph\u1ea3i: \u00c1p d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $a^2-b^2$ ","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$a-1=(\\sqrt{a})^2-1\\,$$=(\\sqrt{a}-1)(\\sqrt{a}+1)$ <br\/><\/span> "}]}],"id_ques":626},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"T\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c $A=\\sqrt{-9x}-\\sqrt{9+12x+4x^2}$ t\u1ea1i $x= -4$ <br\/>\u0110\u00e1p s\u1ed1: $A=$_input_","hint":"\u00c1p d\u1ee5ng: Quy t\u1eafc \u0111\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n ","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 x\u00e1c \u0111\u1ecbnh v\u00e0 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $A$<br\/>B\u01b0\u1edbc 2: Thay $x=-4$ v\u00e0o bi\u1ec3u th\u1ee9c $A$ \u0111\u00e3 r\u00fat g\u1ecdn. <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\le 0$<br\/>Ta c\u00f3:<br\/>$\\begin{align} A&=\\sqrt{-9x}-\\sqrt{9+12x+4x^2}\\\\&=3\\sqrt{-x}-\\sqrt{(2x+3)^2}\\\\&=3\\sqrt{-x}-|3+2x|\\end{align} $<br\/> Thay $x= -4$ v\u00e0o bi\u1ec3u th\u1ee9c $A$, ta \u0111\u01b0\u1ee3c: <br\/> $A= 3\\sqrt{4}-|3-8|=6-5=1$<br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1$ <\/span><\/span>"}]}],"id_ques":627},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["7"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $3\\sqrt{7x}+2\\sqrt{7x}=35$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0: $S=${_input_}","hint":"C\u1ed9ng tr\u1eeb c\u00e1c c\u0103n th\u1ee9c \u0111\u1ed3ng d\u1ea1ng ","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 x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh<br\/>B\u01b0\u1edbc 2: Bi\u1ebfn \u0111\u1ed5i t\u01b0\u01a1ng \u0111\u01b0\u01a1ng v\u00e0 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh.<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh $x\\ge 0$ <br\/>$\\begin{align}& 3\\sqrt{7x}+2\\sqrt{7x}=35\\\\&\\Leftrightarrow 5\\sqrt{7x}=35\\\\&\\Leftrightarrow \\sqrt{7x}=7\\\\&\\Leftrightarrow 7x=49\\\\&\\Leftrightarrow x= 7\\,\\,\\text{(th\u1ecfa m\u00e3n)}\\end{align} $ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ 7 \\right\\}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $7$ <\/span><\/span>"}]}],"id_ques":628},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $3+\\sqrt{5}$","B. $3+2\\sqrt{5}$","C. $3+3\\sqrt{5}$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai5/lv1/img\/4.jpg' \/><\/center>Th\u1ef1c hi\u1ec7n ph\u00e9p t\u00ednh <br\/> $\\sqrt{14+6\\sqrt{5}}=$?","hint":"Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c trong c\u0103n v\u1ec1 $(A+B)^2$ ","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{align}&\\,\\,\\sqrt{14+6\\sqrt{5}}\\\\&=\\sqrt{9+2.3\\sqrt{5}+(\\sqrt{5})^2}\\\\&=\\sqrt{(3+\\sqrt{5})^2}\\\\&=3+\\sqrt{5}\\\\ \\end{align}$<\/span>"}]}],"id_ques":629},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"select":["A. $\\sqrt{3x}$","B. $-\\sqrt{3x}$","C. $2\\sqrt{3x}$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai5/lv1/img\/2.jpg' \/><\/center>V\u1edbi $x \\ge 0$. R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c.<br\/> $5\\sqrt{3x}+4\\sqrt{3x}-10\\sqrt{3x}=$?","hint":"C\u1ed9ng tr\u1eeb c\u00e1c c\u0103n th\u1ee9c \u0111\u1ed3ng d\u1ea1ng ","explain":"<span class='basic_left'>V\u1edbi $x\\ge 0$, ta c\u00f3:<br\/>$5\\sqrt{3x}+4\\sqrt{3x}-10\\sqrt{3x}\\,$$=(5+4-10)\\sqrt{3x}=-\\sqrt{3x}$<\/span>"}]}],"id_ques":630}],"lesson":{"save":0,"level":1}}