ĐKXĐ: x≠2;x≠3
a) Ta có: \(A=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x-4}{3-x}\)
\(=\frac{2x-9}{x^2-2x-3x+6}-\frac{x+3}{x-2}+\frac{2x-4}{x-3}\)
\(=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\frac{\left(2x-4\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\)
\(=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x^2-9}{\left(x-2\right)\left(x-3\right)}+\frac{2\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{2x-9-x^2+9+2\left(x^2-4x+4\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{2x-9-x^2+9+2x^2-8x+8}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{-6x+8+x^2}{\left(x-2\right)\left(x-3\right)}=\frac{x^2-4x-2x+8}{\left(x-2\right)\left(x-3\right)}=\frac{x\left(x-4\right)-2\left(x-4\right)}{\left(x-2\right)\left(x-3\right)}=\frac{\left(x-4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\frac{x-4}{x-3}\)
\(=\frac{x-3-1}{x-3}=1-\frac{1}{x-3}\)
b) Để A nhận giá trị là số nguyên thì \(1-\frac{1}{x-3}\) nhận giá trị nguyên
⇒\(\frac{1}{x-3}\) nhận giá trị nguyên
\(\Rightarrow1⋮x-3\)
hay \(x-3\inƯ\left(1\right)\)
\(\Rightarrow x-3\in\left\{1;-1\right\}\)
\(\Rightarrow x\in\left\{4;2\right\}\)
mà x=2 là không thỏa mãn đkxđ
nên x=4
Vậy: Khi x=4 thì biểu thức \(A=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x-4}{3-x}\) nhận giá trị là số nguyên