a: Đặt \(A=\frac{x\left(x-1\right)}{x^2-25}-\frac{x+3}{x+5}\)
\(=\frac{x\left(x-1\right)-\left(x+3\right)\left(x-5\right)}{\left.\left(x-5\right)\left(x+5\right)\right.}\)
\(=\frac{x^2-x-\left(x^2-2x-15\right)}{\left(x-5\right)\left(x+5\right)}=\frac{3x+15}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{3\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\frac{3}{x-5}\)
Khi x=3 thì \(A=\frac{3}{3-5}=\frac{3}{-2}=-\frac32\)
b: Đặt \(B=\frac{x+3}{x^2-1}-\frac{1}{x^2+x}\)
\(=\frac{x+3}{\left(x-1\right)\left(x+1\right)}-\frac{1}{x\cdot\left(x+1\right)}\)
\(=\frac{x\left(x+3\right)-\left(x-1\right)}{x\left(x+1\right)\left(x-1\right)}=\frac{x^2+2x+1}{x\left(x+1\right)\left(x-1\right)}\)
\(=\frac{\left(x+1\right)^2}{x\left(x+1\right)\left(x-1\right)}=\frac{x+1}{x\left(x-1\right)}\)
Khi x=5 thì \(B=\frac{5+1}{5\left(5-1\right)}=\frac{6}{5\cdot4}=\frac{6}{20}=\frac{3}{10}\)
