{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"select":["A. $\\sqrt{5}$","B. $-3\\sqrt{5}$","C. $3\\sqrt{5}$"],"ques":"V\u1edbi $a > 0$. R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c:<br\/> $\\dfrac{1}{2}\\sqrt{12a}-2\\sqrt{45}-\\dfrac{a}{3}\\sqrt{\\dfrac{27}{a}}+5\\sqrt{1\\dfrac{4}{5}}=$?","hint":"\u00c1p d\u1ee5ng quy t\u1eafc \u0111\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n v\u00e0 tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu ","explain":"<span class='basic_left'>V\u1edbi $a > 0$, ta c\u00f3:<br\/>$\\dfrac{1}{2}\\sqrt{12a}-2\\sqrt{45}-\\dfrac{a}{3}\\sqrt{\\dfrac{27}{a}}+5\\sqrt{1\\dfrac{4}{5}} $<br\/>$=\\dfrac{1}{2}\\sqrt{4.3.a}-2.\\sqrt{9.5}-\\dfrac{a}{3}.\\sqrt{\\dfrac{9.3.a}{{{a}^{2}}}}+5\\sqrt{\\dfrac{9.5}{25}}$<br\/>$ =\\dfrac{2}{2}\\sqrt{3a}-2.3.\\sqrt{5}-\\dfrac{a}{3}.\\dfrac{3}{a}\\sqrt{3a}\\,$$+5.\\dfrac{3}{5}\\sqrt{5} $<br\/>$ =\\sqrt{3a}-6\\sqrt{5}-\\sqrt{3a}+3\\sqrt{5} $<br\/>$ =-3\\sqrt{5} $<\/span>"}]}],"id_ques":631},{"time":4,"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":[[["12"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai5/lv2/img\/4.jpg' \/><\/center><span class='basic_left'>V\u1edbi $x>0;\\,\\,x\\ne 4$. Cho hai bi\u1ec3u th\u1ee9c:<br\/> $P=\\dfrac{x+3}{\\sqrt{x}-2}$v\u00e0 $Q=\\dfrac{\\sqrt{x}-1}{\\sqrt{x}+2}+\\dfrac{5\\sqrt{x}-2}{x-4}$ <br\/><b> C\u00e2u 1: <\/b> V\u1edbi $x= 9$ th\u00ec $P=$_input_<\/span>","hint":"","explain":"<span class='basic_left'>Thay $x= 9$ v\u00e0o bi\u1ec3u th\u1ee9c $P$ ta c\u00f3: <br\/> $P=\\dfrac{9+3}{\\sqrt{9}-2}=\\dfrac{12}{3-2}=12$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $12$ <\/span>"}]}],"id_ques":632},{"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{x}}{\\sqrt{x}-2}$","B. $\\dfrac{1}{\\sqrt{x}-2}$","C. $\\frac{2}{\\sqrt{x}-2}$"],"ques":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai5/lv2/img\/4.jpg' \/><\/center><span class='basic_left'>V\u1edbi $x>0;\\,\\,x\\ne 4$. Cho hai bi\u1ec3u th\u1ee9c:<br\/> $P=\\dfrac{x+3}{\\sqrt{x}-2}$v\u00e0 $Q=\\dfrac{\\sqrt{x}-1}{\\sqrt{x}+2}+\\dfrac{5\\sqrt{x}-2}{x-4}$ <br\/>R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $Q$.<br\/>\u0110\u00e1p s\u1ed1: $Q=$?<\/span>","hint":"X\u00e1c \u0111\u1ecbnh m\u1eabu th\u1ee9c chung r\u1ed3i quy \u0111\u1ed3ng v\u00e0 r\u00fat g\u1ecdn $Q$","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ge 0,\\, x\\ne 4$<br\/>Ta c\u00f3:<br\/>$\\begin{align} Q& =\\dfrac{\\sqrt{x}-1}{\\sqrt{x}+2}+\\dfrac{5\\sqrt{x}-2}{x-4}\\\\&=\\dfrac{\\left( \\sqrt{x}-1 \\right)\\left( \\sqrt{x}-2 \\right)+5\\sqrt{x}-2}{x-4}\\\\&=\\dfrac{x-3\\sqrt{x}+2+5\\sqrt{x}-2}{x-4}\\\\&=\\dfrac{x+2\\sqrt{x}}{x-4}\\\\&=\\dfrac{\\sqrt {x}(\\sqrt x+2)}{(\\sqrt x+2)(\\sqrt x-2)}\\\\&=\\dfrac{\\sqrt{x}}{\\sqrt{x}-2}\\\\\\end{align}$ <br\/>"}]}],"id_ques":633},{"time":4,"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":[[["16"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai5/lv2/img\/4.jpg' \/><\/center><span class='basic_left'>V\u1edbi $x>0;\\,\\,x\\ne 4$. Cho hai bi\u1ec3u th\u1ee9c:<br\/> $P=\\dfrac{x+3}{\\sqrt{x}-2}$v\u00e0 $Q=\\dfrac{\\sqrt{x}-1}{\\sqrt{x}+2}+\\dfrac{5\\sqrt{x}-2}{x-4}$ <br\/><b> C\u00e2u 3: <\/b> V\u1edbi $Q=2$ th\u00ec $x=$_input_<\/span>","hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $Q=2$ (s\u1eed d\u1ee5ng k\u1ebft qu\u1ea3 c\u00e2u 2). ","explain":"<span class='basic_left'>The c\u00e2u 2, v\u1edbi $x\\ge 0,\\, x\\ne 4$, ta c\u00f3:<br\/> $Q=\\dfrac{\\sqrt{x}}{\\sqrt{x}-2}$<br\/>Thay $Q=2$ ta c\u00f3: <br\/> $\\begin{align} & \\,\\,\\,\\dfrac{\\sqrt{x}}{\\sqrt{x}-2}=2\\,\\,\\,\\,\\,\\,\\,\\, \\\\ & \\Leftrightarrow \\dfrac{\\sqrt{x}}{\\sqrt{x}-2}=\\dfrac{2\\left( \\sqrt{x}-2 \\right)}{\\sqrt{x}-2} \\\\ & \\Leftrightarrow \\sqrt{x}=2\\sqrt{x}-4 \\\\ & \\Leftrightarrow \\sqrt{x}=4 \\\\ & \\Leftrightarrow x=16 \\,\\,\\text{(th\u1ecfa m\u00e3n)}\\\\ \\end{align}$ <br\/>V\u1eady $Q= 2$ th\u00ec $x = 16$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $16$ <\/span>"}]}],"id_ques":634},{"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}$","B. $2\\sqrt{3}$","C. $3\\sqrt{3}$"],"ques":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai5/lv2/img\/4.jpg' \/><\/center><span class='basic_left'>V\u1edbi $x>0;\\,\\,x\\ne 4$. Cho hai bi\u1ec3u th\u1ee9c:<br\/> $P=\\dfrac{x+3}{\\sqrt{x}-2}$v\u00e0 $Q=\\dfrac{\\sqrt{x}-1}{\\sqrt{x}+2}+\\dfrac{5\\sqrt{x}-2}{x-4}$ <br\/><b> C\u00e2u 4: <\/b>T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac P Q$<br\/>\u0110\u00e1p s\u1ed1: ${{\\left( \\dfrac{P}{Q} \\right)}_{\\min }}=$?<\/span>","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: Bi\u1ebfn \u0111\u1ed5i $\\dfrac{P}{Q}$ v\u1ec1 d\u1ea1ng $f(x)+\\dfrac {a}{f(x)}$ (v\u1edbi $a \\ge 0; f(x) > 0$)<br\/>B\u01b0\u1edbc 2: \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c Cauchy cho $f(x)+\\dfrac {a}{f(x)}$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'>The c\u00e2u 2, v\u1edbi $x > 0,\\, x\\ne 4$, ta c\u00f3:<br\/>$Q=\\dfrac{\\sqrt{x}}{\\sqrt{x}-2}$<br\/>Ta c\u00f3: $\\dfrac{P}{Q}=\\dfrac{x+3}{\\sqrt{x}-2}:\\dfrac{\\sqrt{x}}{\\sqrt{x}-2}=\\dfrac{x+3}{\\sqrt{x}}\\,$$=\\sqrt{x}+\\dfrac{3}{\\sqrt{x}}$<br\/>Ta c\u00f3: $\\sqrt x > 0; \\dfrac{3}{\\sqrt x}>0 \\,\\,\\forall x$<br\/>\u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c Cauchy cho hai s\u1ed1 d\u01b0\u01a1ng $\\sqrt x$ v\u00e0 $\\dfrac{3}{\\sqrt x}$ ta c\u00f3:<br\/>$\\sqrt{x}+\\dfrac{3}{\\sqrt{x}}\\ge 2\\sqrt{\\sqrt{x}.\\dfrac{3}{\\sqrt{x}}}=2\\sqrt{3}$<br\/>$\\Rightarrow{\\left( \\dfrac{P}{Q} \\right)}\\ge 2\\sqrt {3}\\Rightarrow{{\\left( \\dfrac{P}{Q} \\right)}_{\\min }}=2\\sqrt{3}$.<br\/> D\u1ea5u \u2018=\u2019 x\u1ea3y ra khi $\\sqrt{x}=\\dfrac{3}{\\sqrt{x}}\\Leftrightarrow x=3$<br\/>V\u1eady ${{\\left( \\dfrac{P}{Q} \\right)}_{\\min }}=2\\sqrt{3}\\Leftrightarrow x=3$<\/span><\/span>"}]}],"id_ques":635},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"select":["A. $\\sqrt{2x}$","B. $2\\sqrt{2x}$","C. $4\\sqrt{2x}$"],"ques":"V\u1edbi $x > 0$. R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c:<br\/>$\\sqrt{18x}+x\\sqrt{\\dfrac{2}{x}}=$?","hint":"\u00c1p d\u1ee5ng quy t\u1eafc: \u0110\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n v\u00e0 tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu","explain":"<span class='basic_left'>V\u1edbi $x > 0$, ta c\u00f3:<br\/>$\\sqrt{18x}+x\\sqrt{\\dfrac{2}{x}}=3\\sqrt{2x}+\\dfrac{x\\sqrt{2x}}{x}\\,$$=3\\sqrt{2x}+\\sqrt{2x}=4\\sqrt{2x}$"}]}],"id_ques":636},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $-\\sqrt{5}$","B. $2\\sqrt{5}$","C. $-2\\sqrt{5}$"],"ques":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai5/lv2/img\/4.jpg' \/><\/center>R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c sau:<br\/>$5\\sqrt{\\dfrac{1}{5}}-3\\sqrt{5}+\\dfrac{1}{2}\\sqrt{20}=$?","hint":"","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{align} & \\,\\,\\,\\,5\\sqrt{\\dfrac{1}{5}}-3\\sqrt{5}+\\dfrac{1}{2}\\sqrt{20} \\\\ & =\\sqrt{\\dfrac{25}{5}}-3\\sqrt{5}+\\sqrt{\\dfrac{20}{4}} \\\\ & =\\sqrt{5}-3\\sqrt{5}+\\sqrt{5} \\\\ & =-\\sqrt{5} \\\\ \\end{align}$"}]}],"id_ques":637},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"select":["A. $\\dfrac{2}{5}$","B. $\\dfrac{3}{5}$","C. $\\dfrac{4}{5}$"],"ques":" R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c sau:<br\/>$\\dfrac{\\sqrt{48}-\\sqrt{32}}{\\sqrt{75}-\\sqrt{50}}=$?","hint":"","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\dfrac{\\sqrt{48}-\\sqrt{32}}{\\sqrt{75}-\\sqrt{50}}=\\dfrac{\\sqrt{16.3}-\\sqrt{16.2}}{\\sqrt{25.3}-\\sqrt{25.2}}=\\dfrac{4\\sqrt{3}-4\\sqrt{2}}{5\\sqrt{3}-5\\sqrt{2}}\\,$$=\\dfrac{4\\left( \\sqrt{3}-\\sqrt{2} \\right)}{5\\left( \\sqrt{3}-\\sqrt{2} \\right)}=\\dfrac{4}{5}$"}]}],"id_ques":638},{"time":4,"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":" Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $\\dfrac{5\\sqrt{x}-2}{8\\sqrt{x}+2,5}=\\dfrac{2}{7}$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${_input_}","hint":"Chuy\u1ec3n v\u1ebf, quy \u0111\u1ed3ng v\u00e0 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> \u0110\u1eb7t \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 ngh\u0129a <br\/><b>B\u01b0\u1edbc 2:<\/b> 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\/><b>B\u01b0\u1edbc 3:<\/b> 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\/>Ta c\u00f3:<br\/>$\\dfrac{5\\sqrt{x}-2}{8\\sqrt{x}+2,5}=\\dfrac{2}{7}$ <br\/>$ \\Leftrightarrow \\dfrac{7\\left( 5\\sqrt{x}-2 \\right)}{7\\left( 8\\sqrt{x}+2,5 \\right)}\\,$$-\\dfrac{2\\left( 8\\sqrt{x}+2,5 \\right)}{7\\left( 8\\sqrt{x}+2,5 \\right)}=0 $<br\/>$\\Rightarrow 35\\sqrt{x}-14-16\\sqrt{x}-5=0$<br\/>$\\Leftrightarrow 19\\sqrt{x}=19$<br\/>$ \\Leftrightarrow \\sqrt{x}=1$<br\/>$\\Leftrightarrow x=1$ (th\u1ecfa m\u00e3n)<br\/> V\u1eady 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>"}]}],"id_ques":639},{"time":4,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["6"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":" T\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c $\\sqrt{11-6\\sqrt{2}}+\\sqrt{11+6\\sqrt{2}}=$_input_","hint":"\u0110\u01b0a bi\u1ec3u th\u1ee9c trong c\u0103n v\u1ec1 d\u1ea1ng $(A \\pm B)^2$","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$ \\,\\,\\,\\,\\,\\sqrt{11-6\\sqrt{2}}+\\sqrt{11+6\\sqrt{2}}$<br\/>$=\\sqrt{{{3}^{2}}-2.3.\\sqrt{2}+(\\sqrt{2})^{2}}+\\sqrt{{{3}^{2}}+2.3.\\sqrt{2}+(\\sqrt{2})^{2}}$<br\/>$ =\\sqrt{{{\\left( 3-\\sqrt{2} \\right)}^{2}}}+\\sqrt{{{\\left( 3+\\sqrt{2} \\right)}^{2}}}$<br\/>$ = 3-\\sqrt{2} +3+\\sqrt{2} \\,\\,\\,\\,(\\text {V\u00ec}\\,\\,\\,3 >\\sqrt{2})$<br\/> $=6 $ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $6$ <\/span><\/span>"}]}],"id_ques":640},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"select":["A. $3\\sqrt{5}$","B. $4\\sqrt{5}$","C. $5\\sqrt{5}$"],"ques":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai5/lv2/img\/2.jpg' \/><\/center>R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c sau: <br\/> $\\sqrt{20}-\\sqrt{45}+\\sqrt{125}$=?","hint":"","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{align}&\\sqrt{20}-\\sqrt{45}+\\sqrt{125}\\\\ &=\\sqrt{4.5}-\\sqrt{9.5}+\\sqrt{25.5}\\\\&=2\\sqrt{5}-3\\sqrt{5}+5\\sqrt{5}\\\\&=4\\sqrt{5}\\\\ \\end{align}$"}]}],"id_ques":641},{"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}{\\sqrt{a}+1}$","B. $\\dfrac{-1}{\\sqrt{a}-1}$","C. $\\dfrac{1}{\\sqrt{a}+1}$"],"ques":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai5/lv2/img\/9.jpg' \/><\/center><span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $A=\\dfrac{1}{2\\sqrt{a}-2}-\\dfrac{1}{2\\sqrt{a}+2}+\\dfrac{\\sqrt{a}}{1-a}$<br\/><b> C\u00e2u 1: <\/b> R\u00fat g\u1ecdn $A$<br\/>\u0110\u00e1p s\u1ed1: $A=$?<\/span>","hint":"M\u1eabu th\u1ee9c chung: $(\\sqrt{a}-1)(\\sqrt{a}+1)$ ","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\/><span class='basic_left'><br\/>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $a\\ge 0;\\,\\,a\\ne 1$<br\/>Ta c\u00f3:<br\/>$A=\\dfrac{1}{2\\sqrt{a}-2}-\\dfrac{1}{2\\sqrt{a}+2}+\\dfrac{\\sqrt{a}}{1-a}$ <br\/>$A=\\dfrac{1}{2\\left( \\sqrt{a}-1 \\right)}-\\dfrac{1}{2\\left( \\sqrt{a}+1 \\right)}-\\dfrac{\\sqrt{a}}{a-1}$<br\/>$A=\\dfrac{\\sqrt{a}+1-\\sqrt{a}+1-2\\sqrt{a}}{2(a-1)}$<br\/>$A=\\dfrac{2-2\\sqrt{a}}{-2\\left( 1-a \\right)}$<br\/>$A=\\dfrac{2\\left( 1-\\sqrt{a} \\right)}{-2\\left( 1-a \\right)}$<br\/>$A=\\dfrac{-1}{\\sqrt{a}+1}$ <\/span><\/span>"}]}],"id_ques":642},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"select":["A. $-\\dfrac{1}{5}$","B. $-\\dfrac{2}{5}$","C. $-\\dfrac{3}{5}$"],"ques":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai5/lv2/img\/9.jpg' \/><\/center><span class='basic_left'>Cho bi\u1ec3u th\u1ee9c<br\/>$A=\\dfrac{1}{2\\sqrt{a}-2}-\\dfrac{1}{2\\sqrt{a}+2}+\\dfrac{\\sqrt{a}}{1-a}$<br\/><b> C\u00e2u 2: <\/b>V\u1edbi $a=\\dfrac{4}{9}$ th\u00ec $A=$?<\/span>","hint":"Thay $a=\\dfrac{4}{9}$ v\u00e0o bi\u1ec3u th\u1ee9c $A$ \u0111\u00e3 r\u00fat g\u1ecdn. ","explain":"<span class='basic_left'>Theo c\u00e2u 1, v\u1edbi $a\\ge 0;\\,\\,a\\ne 1$, ta c\u00f3;<br\/> $A=\\dfrac{-1}{\\sqrt{a}+1}$<br\/>Thay $a=\\dfrac{4}{9}$ v\u00e0o bi\u1ec3u th\u1ee9c $A$ ta c\u00f3: <br\/>$A=\\dfrac{-1}{\\sqrt{\\dfrac{4}{9}}+1}=\\dfrac{-1}{\\dfrac{2}{3}+1}=\\dfrac{-3}{5}$<\/span>"}]}],"id_ques":643},{"time":4,"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"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai5/lv2/img\/9.jpg' \/><\/center><span class='basic_left'>Cho bi\u1ec3u th\u1ee9c<br\/>$A=\\dfrac{1}{2\\sqrt{a}-2}-\\dfrac{1}{2\\sqrt{a}+2}+\\dfrac{\\sqrt{a}}{1-a}$<br\/><b> C\u00e2u 3: <\/b>V\u1edbi $\\left| A \\right|=\\dfrac{1}{3}$ th\u00ec $a=$_input_<\/span>","hint":"\u00c1p d\u1ee5ng: $\\left| A \\right|=\\left[ \\begin{align}& A \\,\\,\\, (A\\ge 0)\\\\ & -A \\,\\,(A<0)\\\\ \\end{align} \\right. $","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: \u0110\u00e1nh gi\u00e1 gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $A$ <br\/>B\u01b0\u1edbc 2: Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $A=\\dfrac{1}{3}$ <br\/>B\u01b0\u1edbc 3: K\u1ebft h\u1ee3p v\u1edbi \u0111i\u1ec1u ki\u1ec7n c\u1ee7a b\u00e0i to\u00e1n \u0111\u1ec3 t\u00ecm $a$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Theo c\u00e2u 1, v\u1edbi $a\\ge 0;\\,\\,a\\ne 1$, ta c\u00f3;<br\/> $A=\\dfrac{-1}{\\sqrt{a}+1}$<br\/>V\u00ec $\\sqrt a \\ge 0 \\,\\,\\, \\forall a\\ge 0\\Rightarrow A=\\dfrac{-1}{\\sqrt{a}+1}<0\\,\\,\\forall x \\ge 0$<br\/>Suy ra:$\\left| A \\right|=\\dfrac{1}{3}$<br\/>$\\Leftrightarrow -A=\\dfrac{1}{3}$<br\/>$\\Leftrightarrow\\dfrac{1}{\\sqrt{a}+1}=\\dfrac{1}{3}$<br\/>$\\Rightarrow \\sqrt a +1=3$<br\/>$\\Leftrightarrow \\sqrt a=2$<br\/>$\\Leftrightarrow a=4$ <br\/>V\u1eady v\u1edbi $a= 4\\,$ th\u00ec $\\left| A \\right|=\\dfrac{1}{3}$ <span class='basic_pink'><br\/> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $4$<\/span><\/span>"}]}],"id_ques":644},{"time":4,"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":[[["0"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai5/lv2/img\/9.jpg' \/><\/center><span class='basic_left'>Cho bi\u1ec3u th\u1ee9c<br\/>$A=\\dfrac{1}{2\\sqrt{a}-2}-\\dfrac{1}{2\\sqrt{a}+2}+\\dfrac{\\sqrt{a}}{1-a}$<br\/><b> C\u00e2u 4: <\/b> T\u00ecm $a\\in \\mathbb Z$ \u0111\u1ec3 $A\\in \\mathbb Z$.<br\/>\u0110\u00e1p s\u1ed1: $a=$_input_<\/span>","hint":"$A=\\dfrac{-1}{\\sqrt{a}+1}\\in\\mathbb Z \\Leftrightarrow {\\sqrt{a}+1}$ thu\u1ed9c \u01b0\u1edbc c\u1ee7a $-1$ ","explain":"<span class='basic_left'>Theo c\u00e2u 1, v\u1edbi $a\\ge 0;\\,\\,a\\ne 1$, ta c\u00f3:<br\/> $A=\\dfrac{-1}{\\sqrt{a}+1}$<br\/>\u0110\u1ec3 $A\\in \\mathbb Z\\Leftrightarrow \\dfrac{-1}{\\sqrt{a}+1}\\in \\mathbb Z$<br\/>$\\Leftrightarrow \\sqrt{a}+1$ thu\u1ed9c $\u01af (-1)=\\{1 ;-1\\}$<br\/>Ta c\u00f3 b\u1ea3ng:<br\/><table> <tr> <th>$\\sqrt{a}+1$<\/th> <th>$-1$<\/th> <th>$1$<\/th> <\/tr> <tr> <td>$\\sqrt{a}$<\/td> <td>$-2$<\/td> <td>$0$<\/td> <\/tr> <tr> <td>$a$<\/td> <td>lo\u1ea1i<\/td> <td>$0$<\/td> <\/tr><\/table> <span class='basic_pink'><br\/> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0$<\/span><br\/><span class='basic_left<b>Sai l\u1ea7m c\u1ea7n tr\u00e1nh:<\/b> H\u1ecdc sinh th\u01b0\u1eddng hay qu\u00ean k\u1ebft h\u1ee3p v\u1edbi \u0111i\u1ec1u ki\u1ec7n ban \u0111\u1ea7u c\u1ee7a b\u00e0i to\u00e1n \u0111\u1ec3 lo\u1ea1i nghi\u1ec7m<\/span>"}]}],"id_ques":645},{"time":4,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["48"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","type_check":"","ques":" Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $\\sqrt{48x}-\\sqrt{\\dfrac{75x}{4}}+\\sqrt{\\dfrac{x}{3}}-5\\sqrt{\\dfrac{x}{12}}=12$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${_input_}","hint":"","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 ph\u01b0\u01a1ng tr\u00ecnh. <br\/>B\u01b0\u1edbc 2: \u00c1p d\u1ee5ng quy t\u1eafc: \u0110\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n v\u00e0 tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu<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\/><span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x \\ge 0$<br\/>Ta c\u00f3:<br\/>$\\sqrt{48x}-\\sqrt{\\dfrac{75x}{4}}+\\sqrt{\\dfrac{x}{3}}-5\\sqrt{\\dfrac{x}{12}}=12$<br\/>$\\Leftrightarrow\\sqrt{16.3x}-\\sqrt{\\dfrac{25.3x}{4}}+\\sqrt{\\dfrac{3x}{9}}-5\\sqrt{\\dfrac{3x}{36}}=12$<br\/>$ \\Leftrightarrow 4\\sqrt{3x}-\\dfrac{5}{2}\\sqrt{3x}+\\dfrac{1}{3}\\sqrt{3x}-\\dfrac{5}{6}\\sqrt{3x}=12 $<br\/>$ \\Leftrightarrow \\sqrt{3x}=12 $<br\/>$ \\Leftrightarrow 3x=144 $<br\/>$\\Leftrightarrow x=48$ (th\u1ecfa m\u00e3n)<br\/> V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{48\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $48$ <\/span><\/span><\/span>"}]}],"id_ques":646},{"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{21}}{7}$","B. $\\dfrac{\\sqrt{22}}{7}$","C. $\\dfrac{\\sqrt{23}}{7}$"],"ques":" R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c sau:<br\/>$\\dfrac{\\sqrt{15}-\\sqrt{6}}{\\sqrt{35}-\\sqrt{14}}=$?","hint":"Ph\u00e2n t\u00edch t\u1eed th\u1ee9c v\u00e0 m\u1eabu th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: Ph\u00e2n t\u00edch t\u1eed th\u1ee9c v\u00e0 m\u1eabu th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed. <br\/>B\u01b0\u1edbc 2: R\u00fat g\u1ecdn <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\/>$\\begin{align}&\\,\\,\\,\\dfrac{\\sqrt{15}-\\sqrt{6}}{\\sqrt{35}-\\sqrt{14}}\\\\&=\\dfrac{\\sqrt{3}\\left( \\sqrt{5}-\\sqrt{2} \\right)}{\\sqrt{7}\\left( \\sqrt{5}-\\sqrt{2} \\right)}\\\\&=\\dfrac{\\sqrt{3}}{\\sqrt{7}}\\\\&=\\dfrac{\\sqrt{21}}{7}\\\\ \\end{align}$<\/span>"}]}],"id_ques":647},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"select":["A. $\\sqrt{2}$","B. $2\\sqrt{2}$","C. $3\\sqrt{2}$"],"ques":" R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c sau:<br\/>$\\left( 1+\\sqrt{2}+\\sqrt{3} \\right)\\left( 1+\\sqrt{2}-\\sqrt{3} \\right)=$?","hint":"\u00c1p d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $a^2-b^2$ ","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{align} & \\,\\,\\,\\,\\left( 1+\\sqrt{2}+\\sqrt{3} \\right)\\left( 1+\\sqrt{2}-\\sqrt{3} \\right) \\\\ & ={{\\left( 1+\\sqrt{2} \\right)}^{2}}-(\\sqrt{3})^{2} \\\\ & =1+2\\sqrt{2}+2-3 \\\\ & =2\\sqrt{2} \\\\ \\end{align}$"}]}],"id_ques":648},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"select":["A. $-\\sqrt{5}$","B. $\\sqrt{5}$","C. $-2\\sqrt{5}$"],"ques":" T\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c <br\/>$\\sqrt{14-6\\sqrt{5}}-\\sqrt{14+6\\sqrt{5}}=$?","hint":"Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c trong c\u0103n v\u1ec1 d\u1ea1ng: $(A \\pm B)^2$","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$ \\,\\,\\,\\,\\,\\sqrt{14-6\\sqrt{5}}-\\sqrt{14+6\\sqrt{5}} $<br\/>$=\\sqrt{3^2-2.3.\\sqrt{5}+(\\sqrt{5})^2}-\\sqrt{3^2+2.3.\\sqrt{5}+(\\sqrt{5})^2}$<br\/>$=\\sqrt{{{\\left( 3-\\sqrt{5} \\right)}^{2}}}-\\sqrt{{{\\left( 3+\\sqrt{5} \\right)}^{2}}}$<br\/>$ =\\left| 3-\\sqrt{5} \\right|-3-\\sqrt{5} $<br\/>$ =3-\\sqrt{5}-3-\\sqrt{5}\\,\\,\\,\\,(V\u00ec\\,\\,\\,3 >\\sqrt{5})$<br\/>$=-2\\sqrt{5} $<br\/><i>Ghi nh\u1edb: <\/i>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}\\,\\,\\,\\,\\,\\,\\,\\,\\ge {0} \\\\ & -A\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,\\,\\,\\,\\,\\,\\,{A}<{0} \\\\\\end{align} \\right.$<\/span>"}]}],"id_ques":649},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $-21\\sqrt{a}$","B. $21\\sqrt{a}$","C. $\\sqrt{a}$"],"ques":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai5/lv2/img\/2.jpg' \/><\/center>V\u1edbi $a > 0$. R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c sau: <br\/>$2\\sqrt{a}-\\dfrac{5}{a}\\sqrt{9{{a}^{3}}}+a\\sqrt{\\dfrac{4}{a}}-\\dfrac{2}{{{a}^{2}}}\\sqrt{25{{a}^{5}}} =$?","hint":"\u00c1p d\u1ee5ng quy t\u1eafc: \u0110\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n v\u00e0 tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu ","explain":"<span class='basic_left'> V\u1edbi $x > 0$, ta c\u00f3:<br\/>$2\\sqrt{a}-\\dfrac{5}{a}\\sqrt{9{{a}^{3}}}+a\\sqrt{\\dfrac{4}{a}}-\\dfrac{2}{{{a}^{2}}}\\sqrt{25{{a}^{5}}}$ <br\/>$ =2\\sqrt{a}-\\dfrac{5}{a}\\sqrt{9{{a}^{2}}a}+a.\\sqrt{\\dfrac{4a}{{{a}^{2}}}}\\,$$-\\dfrac{2}{{{a}^{2}}}\\sqrt{25{{a}^{4}}a} $<br\/>$ =2\\sqrt{a}-\\dfrac{5}{a}.3a.\\sqrt{a}+a.\\dfrac{2}{a}\\sqrt{a}\\,$$-\\dfrac{10{{a}^{2}}}{{{a}^{2}}}\\sqrt{a} $<br\/>$=2\\sqrt{a}-15\\sqrt{a}+2\\sqrt{a}-10\\sqrt{a}$<br\/>$=-21\\sqrt{a} $.<\/span>"}]}],"id_ques":650}],"lesson":{"save":0,"level":2}}