cho biểu thức M=\(\left(\frac{\sqrt{x}}{x-36}-\frac{\sqrt{x}-6}{x+6\sqrt{x}}\right):\frac{2\sqrt{x}-6}{x+6\sqrt{x}}\)
rút gọn M
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24 tháng 5 2021
a, \(M=\frac{\sqrt{x}}{\sqrt{x}+6}+\frac{1}{\sqrt{x}-6}+\frac{17\sqrt{x}+30}{\left(\sqrt{x}+6\right)\left(\sqrt{x}-6\right)}\)
\(=\frac{x-6\sqrt{x}+\sqrt{x}+6+17\sqrt{x}+30}{\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}=\frac{12\sqrt{x}+x+36}{\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}=\frac{\sqrt{x}+6}{\sqrt{x}-6}\)
b, Ta có : \(L=N.M\Rightarrow L=\frac{\sqrt{x}+6}{\sqrt{x}-6}.\frac{24}{\sqrt{x}+6}=\frac{24}{\sqrt{x}+6}\)
Vì \(\sqrt{x}+6\ge6\)
\(\Rightarrow\frac{24}{\sqrt{x}+6}\le\frac{24}{6}=4\)
Dấu ''='' xảy ra khi \(\sqrt{x}+6=6\Leftrightarrow x=0\)
Vậy GTLN L là 4 khi x = 0
23 tháng 5 2021
Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
PT
24 tháng 7 2017
a) ĐKXĐ: \(x\ne4\)và \(x>0\)
............................
\(\Leftrightarrow A=\left(\frac{x}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{6}{3\left(\sqrt{x}-2\right)}+\frac{1}{\sqrt{x}+2}\right)\)\(:\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}+\frac{10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\frac{3x-6\sqrt{x}\left(\sqrt{x}+2\right)+3\sqrt{x}\left(\sqrt{x}-2\right)}{3\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x-2}\right)}:\left(\frac{x-2+10-x}{\sqrt{x}-2}\right)\)
\(\Leftrightarrow A=\frac{3x-6x-12\sqrt{x}+3x-6\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\left(\frac{8}{\sqrt{x}-2}\right)\)
\(\Leftrightarrow A=\frac{-18\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\frac{\sqrt{x}-2}{8}\)
\(\Leftrightarrow A=\frac{-3}{4\left(\sqrt{x}+2\right)}\)
Vậy \(A=\frac{-3}{4\left(\sqrt{x}-2\right)}\)với \(x>0\)và \(x\ne4\)
b)Ta có \(A< 2\Leftrightarrow\frac{-3}{4\left(\sqrt{x}-2\right)}< 2\)
\(\Leftrightarrow\frac{-3}{4\left(\sqrt{x}-2\right)}-2< 0\)
\(\Leftrightarrow\frac{-3-8\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-2\right)}< 0\)
\(\Leftrightarrow\frac{-3-8\sqrt{x}-16}{4\left(\sqrt{x}-2\right)}< 0\)
\(\Leftrightarrow\frac{-18-8\sqrt{x}}{4\left(\sqrt{x}-2\right)}< 0\)
\(\Leftrightarrow-18-8\sqrt{x}< 0\)( Vì \(4\left(\sqrt{x}-2\right)>0\)với \(\forall x\))
\(\Leftrightarrow\sqrt{x}< \frac{-9}{4}\)(Vô Nghiệm)
Vậy không có gtr nào của x thỏa mãn A<2
T
5 tháng 8 2017
Điều kiện : \(x\ge0;x\ne4;x\ne9\)
\(A=\left(\frac{1}{1+\sqrt{x}}\right):\left[\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-2\sqrt{x}-3\sqrt{x}+6}\right]\)
\(A=\frac{1}{1+\sqrt{x}}:\left[\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
\(A=\frac{1}{1+\sqrt{x}}:\left[\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
\(A=\frac{1}{1+\sqrt{x}}:\left[\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
\(A=\frac{1}{1+\sqrt{x}}:\left[\frac{x-9-\left(x-4\right)+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
\(A=\frac{1}{1+\sqrt{x}}:\left[\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
\(A=\frac{1}{1+\sqrt{x}}:\frac{1}{\sqrt{x}-2}=\frac{\sqrt{x}-2}{1+\sqrt{x}}\)
20 tháng 9 2019
A=(x+x+yy−xy):(xy+yx+xy−xy−xyx+y)
=\frac{x+\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\frac{x\left(\sqrt{xy}-x\right)\sqrt{xy}+y\left(\sqrt{xy}+y\right)\sqrt{xy}-\left(x+y\right)\left(\sqrt{xy}+y\right)\left(\sqrt{xy}-x\right)}{\sqrt{xy}\left(\sqrt{xy}+y\right)\left(\sqrt{xy}-x\right)}=x+yx+xy+y−xy:xy(xy+y)(xy−x)x(xy−x)xy+y(xy+y)xy−(x+y)(xy+y)(xy−x)
=\frac{x+y}{\sqrt{x}+\sqrt{y}}:\frac{x^2y-x^2\sqrt{xy}+xy^2+y^2\sqrt{xy}-y^2\sqrt{xy}+x^2\sqrt{xy}}{xy^2-x^2y}=x+yx+y:xy2−x2yx2y−x2xy+xy2+y2xy−y2xy+x2xy
=\frac{x+y}{\sqrt{x}+\sqrt{y}}.\frac{xy^2-x^2y}{xy^2+x^2y}=x+yx+y.xy2+x2yxy2−x2y
=\frac{x+y}{\sqrt{x}+\sqrt{y}}.\frac{xy\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)}{xy\left(x+y\right)}=x+yx+y.xy(x+y)xy(y−x)(x+y)
=\sqrt{y}-\sqrt{x}=y−x
6 tháng 7 2019
\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}-1\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}\ne1\end{cases}\Rightarrow}\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}}\)
\(M=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}.\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}+\frac{3\left(\sqrt{x}-1\right)}{x-1}-\frac{6\sqrt{x}-4}{x-1}\)
\(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(b,M< \frac{1}{2}\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}< \frac{1}{2}\)
\(\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{1}{2}< 0\)\(\Rightarrow\frac{2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{2\left(\sqrt{x}+1\right)}< 0\)
\(\Rightarrow\frac{2\sqrt{x}-1-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}< 0\)\(\Rightarrow\frac{\sqrt{x}-2}{2\left(\sqrt{x}+1\right)}< 0\)
Vì \(2\left(\sqrt{x}+1\right)>0\Rightarrow\sqrt{x}-2>0\Rightarrow\sqrt{x}>2\)
\(\Rightarrow\sqrt{x}>\sqrt{4}\Leftrightarrow x>4\)
6 tháng 7 2019
\(M=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\left(x\ge0;x\ne1\right)\)
\(M=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{6\sqrt{x}-4}{x-1}\)
\(M=\frac{x+\sqrt{x}+3\sqrt{x}-3}{\left(\sqrt{x}\right)^2-1^2}-\frac{6\sqrt{x}-4}{x-1}\)
\(M=\frac{x-2\sqrt{x}+1}{x-1}\)
\(M=\frac{\left(\sqrt{x}-1\right)^2}{x-1}\)
\(M=\left(\frac{\sqrt{x}}{x-36}-\frac{\sqrt{x}-6}{x+6\sqrt{x}}\right):\frac{2\sqrt{x}-6}{x+6\sqrt{x}}\)
=\(\left(\frac{\sqrt{x}}{\left(\sqrt{x}\right)^2-6^2}-\frac{\sqrt{x}-6}{\sqrt{x}\left(\sqrt{x}+6\right)}\right):\frac{2\sqrt{x}-6}{\sqrt{x}\left(\sqrt{x}+6\right)}\)
=\(\left(\frac{\sqrt{x}}{\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}-\frac{\sqrt{x}-6}{\sqrt{x}\left(\sqrt{x}+6\right)}\right).\frac{\sqrt{x}\left(\sqrt{x}+6\right)}{2\sqrt{x}-6}\)
=\(\left(\frac{x-\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}{\sqrt{x}\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}\right).\frac{\sqrt{x}\left(\sqrt{x}+6\right)}{2\sqrt{x}-6}\)
=\(\left(\frac{x-x+6\sqrt{x}+6\sqrt{x}-36}{\sqrt{x}\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}\right).\frac{\sqrt{x}\left(\sqrt{x}+6\right)}{2\sqrt{x}-6}\)
=\(\left(\frac{12\sqrt{x}-36}{\sqrt{x}\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}\right).\frac{\sqrt{x}\left(\sqrt{x}+6\right)}{2\sqrt{x}-6}\)
=\(\left(\frac{12\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}\right).\frac{\sqrt{x}\left(\sqrt{x}+6\right)}{2\left(\sqrt{x}-3\right)}\)
=\(\frac{6}{\sqrt{x}-6}\)