Câu hỏi của Hồng Quang

Question

CMR 2m^4 + 2m + 1 >= 0 với mọi m

Despacito · Accepted Answer

ta có $2m^4+2m+1$

$=2m^4-2m^2+\dfrac{1}{2}+2m^2+2m+\dfrac{1}{2}$

$=\left(2m^4-2m^2+\dfrac{1}{2}\right)+\left(2m^2+2m+\dfrac{1}{2}\right)$

$=2\left(m^4-m^2+\dfrac{1}{4}\right)+2\left(m^2+m+\dfrac{1}{4}\right)$

$=2\left(m^4+2.\dfrac{1}{2}m^2+\dfrac{1}{4}\right)+2\left(m^2+2.\dfrac{1}{2}m+\dfrac{1}{4}\right)$

$=2\left(m^2-\dfrac{1}{2}\right)^2+2\left(m+\dfrac{1}{2}\right)^2\ge\forall m$  ( đpcm)Câu trả lời: ta có $2m^4+2m+1$

$=2m^4-2m^2+\dfrac{1}{2}+2m^2+2m+\dfrac{1}{2}$

$=\left(2m^4-2m^2+\dfrac{1}{2}\right)+\left(2m^2+2m+\dfrac{1}{2}\right)$

$=2\left(m^4-m^2+\dfrac{1}{4}\right)+2\left(m^2+m+\dfrac{1}{4}\right)$

$=2\left(m^4+2.\dfrac{1}{2}m^2+\dfrac{1}{4}\right)+2\left(m^2+2.\dfrac{1}{2}m+\dfrac{1}{4}\right)$

$=2\left(m^2-\dfrac{1}{2}\right)^2+2\left(m+\dfrac{1}{2}\right)^2\ge\forall m$  ( đpcm)

Violympic toán 9