Bài rút gọn 

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

\(=\left(x-1\right)-x=x-1-x=-1\left(x>1\right)\)

Bài gpt:

\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}=0\)

Đk:\(-1\le x\le3\)

\(pt\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{\left(x-1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-2}+\sqrt{x-3}\right)=0\)

Dễ thấy:\(\sqrt{x-2}+\sqrt{x-3}=0\) vô nghiệm

Nên \(\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)

Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé

P = \(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)

P = \(\frac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

P = \(\frac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

P = \(\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

P = \(\frac{x-2\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

P = \(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

P = \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

Với \(x=6-2\sqrt{5}=5-2\sqrt{5}+1=\left(\sqrt{5}-1\right)^2\) 

=> P = \(\frac{\sqrt{\left(\sqrt{5}-1\right)^2}+1}{\sqrt{\left(\sqrt{5}-1\right)^2}-3}=\frac{\sqrt{5}-1+1}{\sqrt{5}-1-3}=\frac{\sqrt{5}}{\sqrt{5}-4}=\frac{\sqrt{5}\left(\sqrt{5}+4\right)}{\left(\sqrt{5}-4\right)\left(\sqrt{5}+4\right)}=\frac{5+4\sqrt{5}}{-11}\)

1)

\(M=\frac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{6+4\sqrt{2}}}+\frac{6-4\sqrt{2}}{\sqrt{2}-\sqrt{6-4\sqrt{2}}}\)

\(=\frac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{4+2.2.\sqrt{2}+2}}+\frac{6-4\sqrt{2}}{\sqrt{2}-\sqrt{4-2.2.\sqrt{2}+2}}\)

\(=\frac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{\left(2+\sqrt{2}\right)^2}}+\frac{6-4\sqrt{2}}{\sqrt{2}-\sqrt{\left(2-\sqrt{2}\right)^2}}\)

\(=\frac{6+4\sqrt{2}}{2+2\sqrt{2}}+\frac{6-4\sqrt{2}}{-2+2\sqrt{2}}\)

\(=\frac{2.\left(3+2\sqrt{2}\right)}{2.\left(1+\sqrt{2}\right)}+\frac{2.\left(3-2\sqrt{2}\right)}{2.\left(\sqrt{2}-1\right)}\)

\(=\frac{3+2\sqrt{2}}{\sqrt{2}+1}+\frac{3-2\sqrt{2}}{\sqrt{2}-1}\)

\(=\frac{\left(3+2\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\frac{\left(3-2\sqrt{2}\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}\)

\(=1+\sqrt{2}+\sqrt{2}-1=2\sqrt{2}\)

a

a: \(P=\dfrac{-1+2\sqrt{x}-x+x-2\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}:\dfrac{2x+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

b: Thay \(x=6-2\sqrt{5}\) vào P, ta được:

\(P=\dfrac{\sqrt{5}-1}{\sqrt{5}-2}=3+\sqrt{5}\)

a/ \(P=\frac{1}{\sqrt{xy}}\)

b/ \(x^3=8-6x\)

\(\Rightarrow P=\frac{1}{\sqrt{x\left(x^2+6\right)}}=\frac{1}{\sqrt{x^3+6x}}=\frac{1}{\sqrt{8-6x+6x}}=\frac{1}{2\sqrt{2}}\)