Question

a) Cho A= (\(\dfrac{4x}{2+x}\)+\(\dfrac{8x^2}{4-x^2}\)):(\(\dfrac{x-1}{x^2-2x}\)\(-\)\(\dfrac{2}{x}\))

Rút gọn A. Tìm giá trị của x để A<0

b) CMR: \(\left[\left(n^2+3n+1\right)^2-1\right]\)⋮ 24 với n∈N

Answer 1

a: \(A=\dfrac{4x\left(2-x\right)+8x^2}{\left(2+x\right)\left(2-x\right)}:\dfrac{x-1-2x+4}{x\left(x-2\right)}\)

\(=\dfrac{8x-4x^2+8x^2}{\left(x+2\right)\cdot\left(-1\right)\cdot\left(x-2\right)}\cdot\dfrac{x\left(x-2\right)}{-x+3}\)

\(=\dfrac{8x+4x^2}{\left(x+2\right)\cdot\left(-1\right)}\cdot\dfrac{x}{-x+3}\)

\(=\dfrac{4x\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}\cdot x=\dfrac{4x^2}{x+3}\)

b: \(=\left(n^2+3n+1+1\right)\left(n^2+3n+1-1\right)\)

\(=\left(n^2+3n+2\right)\left(n^2+3n\right)\)

\(=n\left(n+1\right)\left(n+2\right)\left(n+3\right)⋮4!=24\)