{"segment":[{"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":10,"width":40,"content":"","type_input":"","type_check":"","ques":" Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\sqrt{x+8}+\\sqrt{x+3}=5$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${ _input_}","hint":"B\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: \u0110\u1eb7t \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 ngh\u0129a: $\\sqrt{A}$ c\u00f3 ngh\u0129a $\\Leftrightarrow A \\ge 0$<br\/>B\u01b0\u1edbc 2: B\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf<br\/>B\u01b0\u1edbc 3: R\u00fat g\u1ecdn c\u00e1c c\u0103n th\u1ee9c \u0111\u1ed3ng d\u1ea1ng <br\/>B\u01b0\u1edbc 4. Bi\u1ebfn \u0111\u1ed5i ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng $\\sqrt{A}=B\\Leftrightarrow A=B^2$ v\u1edbi $B\\ge 0$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh $\\left\\{ \\begin{aligned} & x+8\\ge 0 \\\\ & x+3\\ge 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\ge -8 \\\\ & x\\ge -3 \\\\ \\end{aligned} \\right.\\Leftrightarrow x\\ge -3$<br\/> Ta c\u00f3: $ \\,\\,\\,\\,\\sqrt{x+8}+\\sqrt{x+3}\\,\\,\\,\\,\\,\\,\\,=5 $<br\/>$\\Leftrightarrow x+8+2\\sqrt{\\left( x+3 \\right)\\left( x+8 \\right)}+x+3=25$<br\/>$\\Leftrightarrow 2\\sqrt{{{x}^{2}}+11x+24}\\,\\,=14-2x $<br\/>$ \\Leftrightarrow \\sqrt{{{x}^{2}}+11x+24}\\,\\,\\,\\,\\,=7-x $<br\/>$\\Leftrightarrow (\\sqrt{{{x}^{2}}+11x+24})^2 = (7 - x)^2$ (V\u1edbi $7-x > 0$)<br\/>$ \\Leftrightarrow {{x}^{2}}+11x+24=49-14x+{{x}^{2}}$<br\/>$ \\Leftrightarrow 25x=25 $<br\/>$ \\Leftrightarrow x\\,\\,\\,\\,\\,\\,=1$ (th\u1ecfa m\u00e3n)<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{1\\}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1$ <\/span><\/span><\/span>"}]}],"id_ques":651},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{1}{3}$","B. $\\dfrac{2}{3}$","C. $\\dfrac{5}{3}$"],"ques":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai5/lv3/img\/2.jpg' \/><\/center>R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c<br\/>$\\left( \\dfrac{1}{\\sqrt{5}-\\sqrt{2}}-\\dfrac{1}{\\sqrt{5}+\\sqrt{2}}+1 \\right).\\dfrac{1}{{{\\left( \\sqrt{2}+1 \\right)}^{2}}}=$?","hint":"Quy \u0111\u1ed3ng c\u00e1c ph\u00e2n th\u1ee9c trong ngo\u1eb7c ","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\left( \\dfrac{1}{\\sqrt{5}-\\sqrt{2}}-\\dfrac{1}{\\sqrt{5}+\\sqrt{2}}+1 \\right).\\dfrac{1}{{{\\left( \\sqrt{2}+1 \\right)}^{2}}} $<br\/>$ =\\left( \\dfrac{\\sqrt{5}+\\sqrt{2}}{5-2}-\\dfrac{\\sqrt{5}-\\sqrt{2}}{5-2}+1 \\right)\\,$$.\\dfrac{1}{{{\\left( \\sqrt{2}+1 \\right)}^{2}}} $<br\/>$=\\left( \\dfrac{\\sqrt{5}+\\sqrt{2}-\\sqrt{5}+\\sqrt 2+3}{5-2} \\right)\\,$$.\\dfrac{1}{{{\\left( \\sqrt{2}+1 \\right)}^{2}}} $<br\/>$ =\\dfrac{3+2\\sqrt{2}}{3}.\\dfrac{1}{3+2\\sqrt{2}}$<br\/>$=\\dfrac{1}{3}$<\/span>"}]}],"id_ques":652},{"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":"V\u1edbi $x> 2$. R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $\\sqrt{x+3+2\\sqrt{x+2}}\\,$$-\\sqrt{x+6-4\\sqrt{x+2}}$","select":["A. $2\\sqrt{x+2}$","B. $\\sqrt{x+2}$","C. $3$","D. $2$"],"hint":"Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c trong c\u0103n v\u1ec1 $(a \\pm b)^2$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c trong c\u0103n v\u1ec1 $(a \\pm b)^2$<br\/>B\u01b0\u1edbc 2: \u00c1p d\u1ee5ng: Quy t\u1eafc \u0111\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n<br\/>B\u01b0\u1edbc 3: R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>V\u1edbi $x > 2$, ta c\u00f3:<br\/> $\\sqrt{x+3+2\\sqrt{x+2}}\\,$$-\\sqrt{x+6-4\\sqrt{x+2}} $<br\/>$=\\sqrt{x+2+2\\sqrt{x+2}+1}\\,$$-\\sqrt{x+2-2.2.\\sqrt{x+2}+{{2}^{2}}} $<br\/>$ =\\sqrt{{{\\left( \\sqrt{x+2}+1 \\right)}^{2}}}-\\sqrt{{{\\left( \\sqrt{x+2}-2 \\right)}^{2}}} $<br\/>$=|\\sqrt{x+2}+1|-|\\sqrt{x+2}-2|$<br\/>$ =\\sqrt{x+2}+1-\\sqrt{x+2}+2 \\,\\,(\\text {V\u00ec}\\,x>2\\Rightarrow \\sqrt{x+2}>\\sqrt 4 \\Leftrightarrow\\sqrt{x+2}>2)$<br\/>$=3 $ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C <\/span><\/span>","column":2}]}],"id_ques":653},{"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/bai5/lv3/img\/4.jpg' \/><\/center>R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $\\dfrac{\\sqrt{5+2\\sqrt{6}}}{\\sqrt{3}-\\sqrt{2}}$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: ","select":["A. $2+5\\sqrt{6}$","B. $5+2\\sqrt{6}$","C. $5$","D. $2$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c trong c\u0103n v\u1ec1 $(a+b)^2$<br\/>B\u01b0\u1edbc 2: \u00c1p d\u1ee5ng: V\u1edbi $A$ l\u00e0 m\u1ed9t bi\u1ec3u th\u1ee9c, ta c\u00f3 $\\sqrt{A^2}=|A|=\\left\\{ \\begin{align} & A\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,\\,\\,\\,\\,\\,\\,{A}\\ge {0} \\\\ & -A\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,\\,\\,\\,\\,\\,\\,{A}< {0} \\\\\\end{align} \\right.$<br\/>B\u01b0\u1edbc 3: Tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/>$\\begin{align}&\\,\\,\\,\\,\\, \\dfrac{\\sqrt{5+2\\sqrt{6}}}{\\sqrt{3}-\\sqrt{2}}\\\\&=\\dfrac{\\sqrt{(\\sqrt{3})^{2}+2\\sqrt{3}.\\sqrt{2}+(\\sqrt{2})^{2}}}{\\sqrt{3}-\\sqrt{2}} \\\\ & =\\dfrac{\\sqrt{{{\\left( \\sqrt{3}+\\sqrt{2} \\right)}^{2}}}}{\\sqrt{3}-\\sqrt{2}} \\\\ & =\\dfrac{\\sqrt{3}+\\sqrt{2}}{\\sqrt{3}-\\sqrt{2}} \\\\ & =\\dfrac{{{\\left( \\sqrt{3}+\\sqrt{2} \\right)}^{2}}}{3-2} \\\\ & =5+2\\sqrt{6} \\\\ \\end{align}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B <\/span><\/span>","column":2}]}],"id_ques":654},{"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":[[["3"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":" R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c<br\/> $\\sqrt{6+2\\sqrt{5}-\\sqrt{29-12\\sqrt{5}}}=$ _input_","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c trong c\u0103n v\u1ec1 $(a-b)^2$<br\/>B\u01b0\u1edbc 2: \u0110\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n. <br\/>B\u01b0\u1edbc 3: R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/>$\\begin{align} & \\,\\,\\,\\,\\sqrt{6+2\\sqrt{5}-\\sqrt{29-12\\sqrt{5}}} \\\\ & =\\sqrt{6+2\\sqrt{5}-\\sqrt{{{\\left( 2\\sqrt{5} \\right)}^{2}}-2.2\\sqrt{5}.3+{{3}^{2}}}} \\\\ & =\\sqrt{6+2\\sqrt{5}-\\sqrt{{{\\left( 2\\sqrt{5}-3 \\right)}^{2}}}} \\\\ & =\\sqrt{6+2\\sqrt{5}-2\\sqrt{5}+3} \\\\ & =\\sqrt{9}\\\\ &=3 \\\\ \\end{align}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $3$ <\/span><\/span> <\/span>"}]}],"id_ques":655},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $A=\\dfrac{2\\sqrt{x}-9}{x-5\\sqrt{x}+6}-\\dfrac{\\sqrt{x}+3}{\\sqrt{x}-2}\\,$$-\\dfrac{2\\sqrt{x}+1}{3-\\sqrt{x}}$ <br\/><b>C\u00e2u 1: <\/b> R\u00fat g\u1ecdn $A$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0:<\/span>","select":["A. $\\dfrac{\\sqrt{x}-1}{\\sqrt{x}-2} $","B. $\\dfrac{\\sqrt{x}+1}{\\sqrt{x}-3} $","C. $\\dfrac{\\sqrt{x}-1}{\\sqrt{x}-3} $ ","D. $\\dfrac{\\sqrt{x}+1}{\\sqrt{x}-2} $"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: X\u00e1c \u0111\u1ecbnh m\u1eabu th\u1ee9c chung v\u00e0 quy \u0111\u1ed3ng <br\/>B\u01b0\u1edbc 2: R\u00fat g\u1ecdn t\u1eed th\u1ee9c<br\/>B\u01b0\u1edbc 3: R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ge 0;\\,\\,x\\ne 4;x\\ne 9$<br\/>Ta c\u00f3:<br\/>$A=\\dfrac{2\\sqrt{x}-9}{x-5\\sqrt{x}+6}-\\dfrac{\\sqrt{x}+3}{\\sqrt{x}-2}-\\dfrac{2\\sqrt{x}+1}{3-\\sqrt{x}}$<br\/>$\\begin{aligned} & =\\dfrac{2\\sqrt{x}-9-\\left( \\sqrt{x}+3 \\right)\\left( \\sqrt{x}-3 \\right)+\\left( 2\\sqrt{x}+1 \\right)\\left( \\sqrt{x}-2 \\right)}{\\left( \\sqrt{x}-2 \\right)\\left( \\sqrt{x}-3 \\right)} \\\\ & =\\dfrac{2\\sqrt{x}-9-x+9+2x-3\\sqrt{x}-2}{\\left( \\sqrt{x}-2 \\right)\\left( \\sqrt{x}-3 \\right)}\\\\&=\\dfrac{x-\\sqrt{x}-2}{\\left( \\sqrt{x}-2 \\right)\\left( \\sqrt{x}-3 \\right)} \\\\& = \\dfrac{x + \\sqrt{x} - 2\\sqrt{x} - 2}{\\left( \\sqrt{x}-2 \\right)\\left( \\sqrt{x}-3 \\right)}\\\\&= \\dfrac{\\sqrt{x}(\\sqrt{x} + 1) - 2(\\sqrt{x} + 1)}{\\left( \\sqrt{x}-2 \\right)\\left( \\sqrt{x}-3 \\right)} \\\\ & =\\dfrac{\\left( \\sqrt{x}+1 \\right)\\left( \\sqrt{x}-2 \\right)}{\\left( \\sqrt{x}-2 \\right)\\left( \\sqrt{x}-3 \\right)}\\\\&=\\dfrac{\\sqrt{x}+1}{\\sqrt{x}-3} \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B <\/span><\/span><\/span>","column":2}]}],"id_ques":656},{"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":[[["-1,9"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'> Cho bi\u1ec3u th\u1ee9c: $A=\\dfrac{2\\sqrt{x}-9}{x-5\\sqrt{x}+6}-\\dfrac{\\sqrt{x}+3}{\\sqrt{x}-2}\\,$$-\\dfrac{2\\sqrt{x}+1}{3-\\sqrt{x}}$<br\/><b> C\u00e2u 2: <\/b> V\u1edbi $x=\\dfrac{2}{3-\\sqrt{5}}$ th\u00ec $A=$_input_ <br\/>(k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t)<\/span>","hint":"Tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu c\u1ee7a gi\u00e1 tr\u1ecb $x$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ednh $\\sqrt{x}$ <br\/>B\u01b0\u1edbc 2: Thay $\\sqrt{x}$ v\u00e0o bi\u1ec3u th\u1ee9c \u0111\u00e3 r\u00fat g\u1ecdn A <br\/>B\u01b0\u1edbc 3: Tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'> Ta c\u00f3:<br\/> $x=\\dfrac{2}{3-\\sqrt{5}}=\\dfrac{2\\left( 3+\\sqrt{5} \\right)}{{{3}^{2}}-(\\sqrt{5})^{2}}\\,$$=\\dfrac{6+2\\sqrt{5}}{4}={{\\left( \\dfrac{\\sqrt{5}+1}{2} \\right)}^{2}}$<br\/>$\\Rightarrow \\sqrt{x}=\\dfrac{\\sqrt{5}+1}{2}$<br\/>Theo c\u00e2u 1, v\u1edbi $x\\ge 0;\\,\\,x\\ne 4;x\\ne 9$, ta c\u00f3: $P=\\dfrac{\\sqrt{x}+1}{\\sqrt{x}-3}$<br\/>Thay $\\sqrt{x}=\\dfrac{\\sqrt{5}+1}{2}$ v\u00e0o bi\u1ec3u th\u1ee9c $P$ ta c\u00f3:<br\/>$P=\\left( \\dfrac{\\sqrt{5}+1}{2}+1 \\right):\\left( \\dfrac{\\sqrt{5}+1}{2}-3 \\right)$<br\/>$\\begin{align}\\,\\,\\,\\,\\,\\,\\,\\,\\,&=\\dfrac{\\sqrt{5}+3}{2}:\\dfrac{\\sqrt{5}-5}{2}\\\\&=\\dfrac{\\sqrt{5}+3}{\\sqrt{5}-5}\\\\&=\\dfrac{\\left( \\sqrt{5}+3 \\right)\\left( \\sqrt{5}+5 \\right)}{(\\sqrt{5})^{2}-{{5}^{2}}}\\\\&=\\dfrac{20+8\\sqrt{5}}{-20}\\\\&=-\\dfrac{5+2\\sqrt{5}}{5}\\approx -1,9 \\\\ \\end {align}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-1,9$ <\/span><\/span><\/span>"}]}],"id_ques":657},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["8"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $A=\\dfrac{2\\sqrt{x}-9}{x-5\\sqrt{x}+6}-\\dfrac{\\sqrt{x}+3}{\\sqrt{x}-2}\\,$$-\\dfrac{2\\sqrt{x}+1}{3-\\sqrt{x}}$<br\/><b> C\u00e2u 3: <\/b> Gi\u00e1 tr\u1ecb nguy\u00ean l\u1edbn nh\u1ea5t c\u1ee7a $x$ \u0111\u1ec3 $P< 1$.<br\/> \u0110\u00e1p s\u1ed1: $x=$_input_<\/span>","hint":"Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh $P<1$","explain":"<span class='basic_left'>Theo c\u00e2u 1, v\u1edbi $x\\ge 0;\\,\\,x\\ne 4;x\\ne 9$, ta c\u00f3: <br\/>$P=\\dfrac{\\sqrt{x}+1}{\\sqrt{x}-3}$<br\/> $P<1\\Leftrightarrow \\dfrac{\\sqrt{x}+1}{\\sqrt{x}-3}<1\\,$$\\Leftrightarrow \\dfrac{\\sqrt{x}+1}{\\sqrt{x}-3}-1<0\\Leftrightarrow \\dfrac{4}{\\sqrt{x}-3}<0$<br\/>$\\Leftrightarrow \\sqrt{x}-3<0\\Leftrightarrow \\sqrt{x}<3\\Leftrightarrow x<9$<br\/>K\u1ebft h\u1ee3p v\u1edbi \u0111i\u1ec1u ki\u1ec7n $x\\ge 0;\\,\\,x\\ne 4$<br\/>V\u1eady $P < 1$ n\u1ebfu $0\\le x\\,<\\,9 ;\\,\\,x\\ne 4$ <br\/> V\u1eady gi\u00e1 tr\u1ecb nguy\u00ean l\u1edbn nh\u1ea5t c\u1ee7a $x$ l\u00e0 $8$<br\/><span class='basic_pink'>V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $8$ <\/span><\/span>"}]}],"id_ques":658},{"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":10,"width":40,"content":"","type_input":"","type_check":"","ques":" <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $A=\\dfrac{2\\sqrt{x}-9}{x-5\\sqrt{x}+6}-\\dfrac{\\sqrt{x}+3}{\\sqrt{x}-2}\\,$$-\\dfrac{2\\sqrt{x}+1}{3-\\sqrt{x}}$<br\/><b> C\u00e2u 4: <\/b> Gi\u00e1 tr\u1ecb nguy\u00ean nh\u1ecf nh\u1ea5t c\u1ee7a $x$ \u0111\u1ec3 $P\\in \\mathbb Z$<br\/> \u0110\u00e1p s\u1ed1: $x=$_input_<\/span>","hint":"Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c v\u1ec1 d\u1ea1ng $m+\\dfrac{n}{f(x)}$ v\u1edbi $m,n \\in \\mathbb Z$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c.<br\/>B\u01b0\u1edbc 2: Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c v\u1ec1 d\u1ea1ng $m+\\dfrac{n}{f(x)}$ v\u1edbi $m,n \\in \\mathbb Z$<br\/>B\u01b0\u1edbc 3: Cho m\u1eabu c\u1ee7a ph\u00e2n th\u1ee9c \u0111\u00f3 b\u1eb1ng \u01b0\u1edbc c\u1ee7a t\u1eed <br\/>B\u01b0\u1edbc 4: T\u00ecm $x$, so s\u00e1nh v\u1edbi \u0111i\u1ec1u ki\u1ec7n v\u00e0 k\u1ebft lu\u1eadn <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'> Theo c\u00e2u 1, v\u1edbi $x\\ge 0;\\,\\,x\\ne 4;x\\ne 9$, ta c\u00f3: <br\/> $P=\\dfrac{\\sqrt{x}+1}{\\sqrt{x}-3}=\\dfrac{\\sqrt{x}-3+4}{\\sqrt{x}-3}\\,$$=1+\\dfrac{4}{\\sqrt{x}-3}$<br\/>$P\\in \\mathbb Z\\Leftrightarrow \\dfrac{4}{\\sqrt{x}-3}\\in Z\\,$$\\Leftrightarrow 4\\,\\,\\vdots \\,\\,\\left( \\sqrt{x}-3 \\right)$<br\/>$\\Leftrightarrow \\sqrt{x}\\,-\\,3$ thu\u1ed9c $\u01af(4)=\\left\\{ \\pm 1;\\pm 2;\\pm 4 \\right\\}$<br\/>Ta c\u00f3 b\u1ea3ng gi\u00e1 tr\u1ecb: <br\/><table> <tr> <th>$\\sqrt{x}-3$<\/th> <th>$-4$<\/th> <th>$-2$<\/th> <th>$-1$<\/th> <th>$1$<\/th> <th>2<\/th> <th>$4$<\/th> <\/tr> <tr> <td>$\\sqrt{x}$<\/td> <td>$-1$<\/td> <td>$1$<\/td> <td>$2$<\/td> <td>$4$<\/td> <td>$5$<\/td> <td>$7$<\/td> <\/tr> <tr> <td>$x$<\/td> <td>lo\u1ea1i<\/td> <td>$1$<\/td> <td>$4$(lo\u1ea1i)<\/td> <td>$16$<\/td> <td>$25$<\/td> <td>$49$<\/td> <\/tr><\/table><br\/>Suy ra gi\u00e1 tr\u1ecb nguy\u00ean nh\u1ecf nh\u1ea5t c\u1ee7a $x$ \u0111\u1ec3 $P$ nguy\u00ean l\u00e0 $x=1$<br\/><\/span> <span class='basic_pink'>V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1$ <\/span><\/span>"}]}],"id_ques":659},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"V\u1edbi $a+b >0$ v\u00e0 $b \\ne 0$. K\u1ebft qu\u1ea3 r\u00fat g\u1ecdn c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{a+b}{{{b}^{2}}}\\sqrt{\\dfrac{{{a}^{2}}{{b}^{4}}}{{{a}^{2}}+2ab+{{b}^{2}}}}$ l\u00e0: ","select":["A. $2a$","B. $-a$","C. $a$ ","D. $|a|$"],"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'>V\u1edbi $a+b >0$ v\u00e0 $b \\ne 0$, ta c\u00f3:<br\/>$\\begin{align} & \\,\\,\\,\\,\\dfrac{a+b}{{{b}^{2}}}\\sqrt{\\dfrac{{{a}^{2}}{{b}^{4}}}{{{a}^{2}}+2ab+{{b}^{2}}}} \\\\ & =\\dfrac{a+b}{{{b}^{2}}}.\\sqrt{\\dfrac{{{\\left( a{{b}^{2}} \\right)}^{2}}}{{{\\left( a+b \\right)}^{2}}}} \\\\ & =\\dfrac{a+b}{{{b}^{2}}}.\\left| \\dfrac{a{{b}^{2}}}{a+b} \\right| \\\\ & =\\dfrac{a+b}{{{b}^{2}}}.\\dfrac{|a|{{b}^{2}}}{a+b}=|a| \\,\\,(V\u00ec\\,\\,a+b> 0)\\\\ \\end{align}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D <\/span> <br\/><span class='basic_left'><span class='basic_green'>Sai l\u1ea7m c\u1ea7n tr\u00e1nh: <\/span> H\u1ecdc sinh th\u01b0\u1eddng qu\u00ean ki\u1ec3m tra xem \u0111i\u1ec1u ki\u1ec7n c\u1ee7a bi\u1ec3u th\u1ee9c \u0111\u01b0a ra ngo\u00e0i c\u0103n c\u00f3 d\u01b0\u01a1ng kh\u00f4ng.<\/span><\/span>","column":2}]}],"id_ques":660}],"lesson":{"save":0,"level":3}}