Question

Chứng minh rằng : x^4+4x+5>0

Answer 1

\(x^4+4x+5\)

\(=\left(x^4+4x+4\right)+1\)

\(=\left(x+2\right)^2+1\)

vì \(\left(x+2\right)^2\ge0\)với mọi x

\(\Rightarrow\left(x+2\right)^2+1>0\)với mọi x

vậy.....(đpcm)

Answer 2

@Cold Wind

x^4 +4x +5 > 0

<=>x^4 -2x^2 +1 +2x^2 +4x +4 =0

<=>(x^2 -1)^2 +2 x^2 +4x +2 +2 =0

<=>(x^2 -1)^2 +2 (x+1)^2 +2 =0

có

(x^2 -1)^2 >=0

2 (x+1)^2 >=0

2 >0 => tổng VT >0 => dpcm