a) Ta có: \(A=\frac{1}{x-2}+\frac{x^2-x-2}{x^2-7x+10}-\frac{2x-4}{x-5}\)
\(=\frac{1}{x-2}+\frac{x^2-2x+x-2}{x^2-5x-2x+10}-\frac{2x-4}{x-5}\)
\(=\frac{1}{x-2}+\frac{x\left(x-2\right)+\left(x-2\right)}{x\left(x-5\right)-2\left(x-5\right)}-\frac{2x-4}{x-5}\)
\(=\frac{1}{x-2}+\frac{\left(x-2\right)\left(x+1\right)}{\left(x-5\right)\left(x-2\right)}-\frac{2x-4}{x-5}\)
\(=\frac{1}{x-2}+\frac{x+1}{x-5}-\frac{2x-4}{x-5}\)
\(=\frac{1}{x-2}+\frac{x+1-2x+4}{x-5}\)
\(\frac{1}{x-2}+\frac{5-x}{x-5}=\frac{1}{x-2}-\frac{x-5}{x-5}=\frac{1}{x-2}-1=\frac{1}{x-2}-\frac{x-2}{x-2}=\frac{1-x+2}{x-2}=\frac{-x+3}{x-2}=\frac{3-x}{x-2}\)
b) Để A có giá trị nguyên thì 3-x⋮x-2
⇔-x+2+1⋮x-2
⇔-(x-2)+1⋮x-2
⇔1⋮x-2
hay x-2∈Ư(1)
⇔x-2∈{1;-1}
hay x∈{3;1}
Vậy: x∈{3;1}
