a: \(A=\frac{\sqrt{x-\sqrt{4\left(x-1\right)}}+\sqrt{x+\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}\cdot\left(1-\frac{1}{x-1}\right)\)
\(=\frac{\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{x^2-4x+4}}\cdot\frac{x-1-1}{x-1}\)
\(=\frac{\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}}{\sqrt{\left(x-2\right)^2}}\cdot\frac{x-2}{x-1}\)
\(=\frac{\left|\sqrt{x-1}-1\right|+\sqrt{x-1}+1}{\left|x-2\right|}\cdot\frac{x-2}{x-1}\)
TH1: x>2
=>x-2>0; \(\sqrt{x-1}-1>0\)
\(A=\frac{\left|\sqrt{x-1}-1\right|+\sqrt{x-1}+1}{\left|x-2\right|}\cdot\frac{x-2}{x-1}\)
\(=\frac{\sqrt{x-1}-1+\sqrt{x-1}+1}{x-2}\cdot\frac{x-2}{x-1}=\frac{2\sqrt{x-1}}{x-1}=\frac{2}{\sqrt{x-1}}\)
TH2: 1<x<2
=>x-2<0; \(\sqrt{x-1}-1<0\)
Ta có: \(A=\frac{\left|\sqrt{x-1}-1\right|+\sqrt{x-1}+1}{\left|x-2\right|}\cdot\frac{x-2}{x-1}\)
\(=\frac{-\sqrt{x-1}+1+\sqrt{x-1}+1}{-\left(x-2\right)}\cdot\frac{x-2}{x-1}=\frac{2}{-1\cdot\left(x-1\right)}=\frac{-2}{x-1}\)
b: TH1: x>2
=>\(A=\frac{2}{\sqrt{x-1}}\)
Để A là số nguyên thì 2⋮\(\sqrt{x-1}\)
=>\(\sqrt{x-1}\in\left\lbrace1;2\right\rbrace\)
=>x-1∈{1;4}
=>x∈{2;5}
=>x=5
TH2: 1<x<2
=>\(A=-\frac{2}{x-1}\)
Để A là số nguyên thì -2⋮x-1
=>x-1∈{1;-1;2;-2}
=>x∈{2;0;3;-1}
mà 1<x<2
nên x∈∅
