Câu a: 

TH1: \(x+\sqrt{\left(2x-1\right)^2}=2\Leftrightarrow x+2x-1=2\Leftrightarrow x=1\)

TH2:\(x+\sqrt{\left(2x-1\right)^2}=2\Leftrightarrow x-2x+1=2\Leftrightarrow x=-1\)

ĐK:  \(x\le2\)

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

\(\Leftrightarrow\)\(\sqrt{4x^2-4x+1}=2-x\)

\(\Leftrightarrow\)\(4x^2-4x+1=4-4x+x^2\)

\(\Leftrightarrow\)\(3x^2=3\)

\(\Leftrightarrow\)\(x=\pm1\)(t/m)

Vậy...

\(1-\sqrt{4x^2-20x+25}=0\)

\(\Leftrightarrow\)\(\sqrt{4x^2-20x+25}=1\)

\(\Leftrightarrow\)\(4x^2-20x+24=0\)

\(\Leftrightarrow\)\(x^2-5x+6=0\)

\(\Leftrightarrow\)\(\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

Vậy...

ok k đi

You 2k mấy mak đòi xưng a -_-

ab + 2 = 2 + ab :))))

P=\(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\)

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

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

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

\(=\frac{\sqrt{x}+1}{\sqrt{x}}=\frac{\sqrt{x}}{\sqrt{x}}+\frac{1}{\sqrt{x}}=1+\frac{1}{\sqrt{x}}\)

Để\(P\in Z\)<=>\(\frac{1}{\sqrt{x}}\in Z\Leftrightarrow\sqrt{x}\inƯ\left(1\right)=1\)\(Với\sqrt{x}=1\Leftrightarrow x=1\)loại

Vậy không có giá trị x nào thỏa mãn P\(\in\)Z