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\(\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{x+\sqrt{x}}-\frac{2}{1-x}\right)\) (ĐKXĐ : \(x>0;x\ne1;x\ne\frac{1}{9}\) )
\(=\left[\frac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\left[\frac{\sqrt{x}-1}{\sqrt{x}.\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}}{\sqrt{x}.\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{3\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{3\sqrt{x}-1}\)
\(=\frac{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}{3\sqrt{x}-1}\)
\(A=\left(\frac{3}{\sqrt{x}-1}-\frac{\sqrt{x}-3}{x-1}\right):\left(\frac{x+2}{x+\sqrt{x}-2}-\frac{\sqrt{x}}{\sqrt{x}+2}\right)\left(ĐK:x\ge0;\ne1\right)\)
\(=\left[\frac{3}{\sqrt{x}-1}-\frac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\left[\frac{x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}}{\sqrt{x}+2}\right]\)
\(=\frac{3\left(\sqrt{x}+1\right)-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{x+2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{2\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\)
\(=\frac{2\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}=\frac{2\left(\sqrt{x}+3\right)}{\sqrt{x}+1}\)
a)ĐKXĐ:x>=0;x khác 9
A=[\(\frac{\sqrt{x}}{\sqrt{x}-3}\) - \(\frac{3\sqrt{x}+9}{x-9}\)+ \(\frac{2\sqrt{x}}{\sqrt{x}+3}\)] \(\div\) [\(\frac{2\sqrt{x}-2}{\sqrt{x}-3}\)-1]
A=[\(\frac{\sqrt{x}\left(\sqrt{x}-3\right)-3\sqrt{x}-9+2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}\)] \(\div\) [\(\frac{\left(2\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-x+9}{x-9}\)]
A=[\(\frac{3x-12\sqrt{x}-9}{x-9}\)].[\(\frac{x-9}{x-4\sqrt{x}+3}\)]
A=\(\frac{3x-12\sqrt{x}-9}{x-4\sqrt{x}+3}\)
Bài 1:
\(=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
Sửa đề; \(D=\left(\dfrac{\sqrt{x}+\sqrt{y}}{2\sqrt{x}-2\sqrt{y}}-\dfrac{2\sqrt{xy}}{x-y}\right)\cdot\dfrac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(D=\dfrac{x+2\sqrt{xy}+y-4\sqrt{xy}}{2\left(x-y\right)}\cdot\dfrac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x}-\sqrt{y}}\cdot\dfrac{\sqrt{x}}{x-y}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)
\(=\dfrac{a\sqrt{a}-3-2\left(a-6\sqrt{a}+9\right)-a-4\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{a-1}{a+8}\)
\(=\dfrac{a\sqrt{a}-a-4\sqrt{a}-6-2a+12\sqrt{a}-18}{\left(\sqrt{a}-3\right)}\cdot\dfrac{\sqrt{a}-1}{a+8}\)
\(=\dfrac{a\sqrt{a}-3a+8\sqrt{a}-24}{\left(\sqrt{a}-3\right)}\cdot\dfrac{\sqrt{a}-1}{a+8}=\sqrt{a}-1\)
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