Question

với mọi số thực x chứng minh \(\left(x-2\right)\left(x+3\right)\left(x+4\right)\left(x-6\right)+57x^2>0\)

Answer 1

- Với \(x=0\Rightarrow144>0\) (đúng)

- Với \(x\ne0\)

\(VT=\left(x-2\right)\left(x-6\right)\left(x+3\right)\left(x+4\right)+57x^2\)

\(=\left(x^2+12-8x\right)\left(x^2+12+7x\right)+57x^2\)

\(=x^2\left[\left(x+\frac{12}{x}-8\right)\left(x+\frac{12}{x}+7\right)+57\right]\)

\(=x^2\left[\left(x+\frac{12}{x}-8\right)^2+15\left(x+\frac{12}{x}-8\right)+57\right]\)

\(=x^2\left[\left(x+\frac{12}{x}-8+\frac{15}{2}\right)^2+\frac{3}{4}\right]>0;\forall x\ne0\)

Vậy...