Question

Chứng minh với x thuộc Z, x khác -1 thì \(\dfrac{x^4+3x^3+3x^2+x}{2x^2+4x+2}\) nguyên.

Answer 1

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

\(=\dfrac{x^3\left(x+1\right)+2x^2\left(x+1\right)+x\left(x+1\right)}{2\left(x+1\right)^2}\)

\(=\dfrac{\left(x+1\right)\left(x^3+2x^2+x\right)}{2\left(x+1\right)^2}\)

\(=\dfrac{x\left(x+1\right)^2\cdot\left(x+1\right)}{2\left(x+1\right)^2}=\dfrac{x\left(x+1\right)}{2}\)

Vì x;x+1 là hai số liên tiếp

nên \(x\left(x+1\right)⋮2\)

=>x(x+1)/2 là số nguyên