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

+) Nếu  x < 1 => A = \(\frac{-\left(x-1\right)}{x-1}-\frac{-\left(x-2\right)}{x-2}=-1-\left(-1\right)=0\)

+) Nếu 1 < x < 2 => A = \(\frac{\left(x-1\right)}{x-1}-\frac{-\left(x-2\right)}{x-2}=1-\left(-1\right)=2\)

+) Nếu x > 2 => A = \(\frac{\left(x-1\right)}{x-1}-\frac{\left(x-2\right)}{x-2}=1-1=0\)

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

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

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

em lớp 8 chỉ làm được thế thôi

\(A=1-\left(\frac{2}{1+2\sqrt{x}}-\frac{5\sqrt{x}}{4x-1}-\frac{1}{1-2\sqrt{x}}\right):\frac{\sqrt{x}-1}{4x+4\sqrt{x}+1}\)

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

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

để A là số nguyên thì \(1-2\sqrt{x}\) là ước của 2 khi đó ta tìm được \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)