Question

cmr đa thức P(x)=x²+x-2017 không thể có nghiệm nguyên

Answer 1

tách thành HĐT

Answer 2

\(P\left(x\right)=x^2+x-2017=x^2+x+1-2018\)

\(P\left(x\right)=x^2+\frac{1}{2}x+\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}-2018\)

\(P\left(x\right)=x\left(x+\frac{1}{2}\right)+\frac{1}{2}\left(x+\frac{1}{2}\right)+\frac{3}{4}-2018\)

\(P\left(x\right)=\left(x+\frac{1}{2}\right)\left(x+\frac{1}{2}\right)+\frac{3}{4}-2018=\left(x+\frac{1}{2}\right)^2-\frac{8069}{4}\)

Vì \(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow\left(x+\frac{1}{2}\right)^2-\frac{8069}{4}\ge\frac{-8069}{4}\)

=>P(x) vô nghiệm