a: \(A=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{x+3}{x-2}+\dfrac{2x+1}{x-3}\)
\(=\dfrac{2x-9-x^2+9+\left(2x+1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\)
\(=\dfrac{-x^2+2x+2x^2-4x+x-2}{\left(x-3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-x-2}{\left(x-3\right)\left(x-2\right)}=\dfrac{x+1}{x-3}\)
b: Ta có: \(\left|2x-1\right|=3\)
=>2x-1=3 hoặc 2x-1=-3
=>2x=4 hoặc 2x=-2
=>x=2(loại) hoặc x=-1(nhận)
Thay x=-1 vào A, ta được:
\(A=\dfrac{-1+1}{-1-3}=0\)
d: Để \(A=\dfrac{x}{x+2}\) thì \(\dfrac{x+1}{x-3}=\dfrac{x}{x+2}\)
\(\Leftrightarrow x^2+3x+2=x^2-3x\)
=>6x=-2
hay x=-1/3(nhận)