Question

Chứng minh rằng: \(x^8-x^5-\dfrac{1}{x}+\dfrac{1}{x^4}\ge0\) với mọi \(x\),\(x\ne0\)

Answer 1

Giải:

\(x^8-x^5-\dfrac{1}{x}+\dfrac{1}{x^4}\ge0\)

\(\Leftrightarrow x^4\left(x^8-x^5-\dfrac{1}{x}+\dfrac{1}{x^4}\right)\ge0\)

\(\Leftrightarrow x^{12}-x^9-x^3+1\ge0\)

\(\Leftrightarrow x^9\left(x^3-1\right)-\left(x^3-1\right)\ge0\)

\(\Leftrightarrow\left(x^3-1\right)\left(x^9-1\right)\ge0\)

\(\Leftrightarrow\left(x^3-1\right)\left(x^3-1\right)\left(x^6+x^3+1\right)\ge0\)

\(\Leftrightarrow\left(x^3-1\right)^2\left(x^6+x^3+1\right)\ge0\) (luôn đúng)

Vậy ...