Bài 5:
a: \(A=\dfrac{2x^2-4x}{x\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x+2\right)}-\dfrac{2}{x-2}\)
\(=\dfrac{2x^2-4x+\left(x^2-4x+4\right)\left(x+2\right)-2x\left(x+2\right)}{x\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{2x^2-4x-2x^2-2x+x^3+2x^2-4x^2-8x+4x+8}{x\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{x^3-2x^2-10x+8}{x\left(x+2\right)\left(x-2\right)}\)
b: Thay x=4 vào A, ta được:
\(A=\dfrac{4^3-2\cdot4^2-10\cdot4+8}{4\left(4+2\right)\left(4-2\right)}=0\)





