Question

Answer 1

a: ĐKXĐ: \(x\notin\left\{0;8\right\}\)

Sửa đề: \(B=\left(\dfrac{4x+2}{x^2-8x}+\dfrac{4x-2}{x^2+8x}\right)\cdot\dfrac{x^2-64}{x^2+4}\)

\(=\left(\dfrac{4x+2}{x\left(x-8\right)}+\dfrac{4x-2}{x\left(x+8\right)}\right)\cdot\dfrac{\left(x-8\right)\left(x+8\right)}{x^2+4}\)

\(=\dfrac{x\left(4x+2\right)+x\left(4x-2\right)}{x\left(x-8\right)\left(x+8\right)}\cdot\dfrac{\left(x-8\right)\left(x+8\right)}{x^2+4}\)

\(=\dfrac{8x^2}{x\left(x^2+4\right)}=\dfrac{8x}{x^2+4}\)

b: Khi x=0 thì \(B=\dfrac{8\cdot0}{0^2+4}=\dfrac{8}{4}=2\)

Khi x=1 thì \(B=\dfrac{8\cdot1}{1^2+4}=\dfrac{8}{5}\)