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30 tháng 8 2021

Tìm đk , rút gọnundefined

30 tháng 8 2021

ĐK : x > 2 

\(\frac{\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{x^2-4\left(x-1\right)}}\left(1-\frac{1}{x-1}\right)\)

\(=\frac{\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}}{\sqrt{x^2-4x+4}}\left(\frac{x-1-1}{x-1}\right)\)

\(=\frac{\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}}{\sqrt{\left(x-2\right)^2}}\left(\frac{x-2}{x-1}\right)\)

Với x > 2 

\(=\frac{\sqrt{x-1}-1+\sqrt{x-1}+1}{x-2}\left(\frac{x-2}{x-1}\right)=\frac{2\sqrt{x-1}}{x-1}\)

7 tháng 7 2015

a)\(x-4\ne0;x\ge0\)

<=>\(x\ne4;x\ge0\)

b)\(B=\left(\frac{1}{\sqrt{x}+2}-\frac{2}{x+4\sqrt{x}+4}\right):\left(\frac{2}{x-4}-\frac{1}{\sqrt{x}-2}\right)\)

=\(\left(\frac{1}{\sqrt{x}+2}-\frac{2}{\left(\sqrt{x}+2\right)^2}\right):\left(\frac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{1}{\sqrt{x}-2}\right)\)

=\(\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2}-\frac{2}{\left(\sqrt{x}+2\right)^2}\right):\left(\frac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)

=\(\frac{\sqrt{x}}{\left(\sqrt{x}+2\right)^2}:\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

=\(\frac{\sqrt{x}}{\left(\sqrt{x}+2\right)^2}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}}\)

=\(\frac{\sqrt{x}-2}{\sqrt{x}+2}\)

a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)

b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)

\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)

c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)

\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)

\(=\dfrac{3}{\sqrt{x}-2}\)