Question

chứng minh 2x² - 3x + 4 > 0 với mọi x

 

 

 

Answer 1

Có: \(2x^2-3x+4=2\left(x^2-\frac{3}{2}x+\frac{9}{16}\right)+\frac{23}{8}=2\left(x-\frac{3}{4}\right)^2+\frac{23}{8}\)

Vì: \(2\left(x-\frac{3}{4}\right)^2\ge0,\forall x\)

=> \(2\left(x-\frac{3}{4}\right)^2+\frac{23}{8}>0,\forall x\)

=>đpcm

Answer 2

\(2x^2-3x+4=2\left(x^2-\frac{3}{2}x+2\right)=2\left(x^2-2.\frac{3}{4}x+\frac{9}{16}+\frac{23}{16}\right)=2\left(x-\frac{3}{4}\right)^2+\frac{23}{8}>0\)

mà \(2\left(x-\frac{3}{4}\right)^2>=0\) => ĐPCM