Bài 1:
a: (x-5)(x+2)+3(x-2)(x+2)-4x2+3x
\(=x^2-3x-10+3\left(x^2-4\right)-4x^2+3x\)
\(=-3x^2-10+3x^2-12\)
=-10-12
=-22
b: \(x^3+3x^2+5x+a⋮x+3\)
=>\(x^3+3x^2+5x+15+a-15⋮x+3\)
=>\(\left(x+3\right)\left(x^2+5\right)+a-15⋮x+3\)
=>a-15=0
=>a=15
Bài 2:
1: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
2: \(A=\left(\dfrac{x+1}{x-1}+\dfrac{8}{x^2-1}-\dfrac{x-1}{x+1}\right)\cdot\dfrac{x-1}{x+1}\)
\(=\left(\dfrac{x+1}{x-1}+\dfrac{8}{\left(x-1\right)\left(x+1\right)}-\dfrac{x-1}{x+1}\right)\cdot\dfrac{x-1}{x+1}\)
\(=\dfrac{\left(x+1\right)^2+8-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x-1}{x+1}\)
\(=\dfrac{x^2+2x+1+8-x^2+2x-1}{x+1}\cdot\dfrac{1}{x+1}\)
\(=\dfrac{4x+8}{\left(x+1\right)^2}\)