a ) \(x^2+6x+10\)
\(=\left(x^2+2.x.3+3^2\right)+1\)
\(=\left(x+3\right)^2+1\ge1>0\) ( đpcm )
b ) \(x^2-x+1\)
\(=\left(x^2-2.\frac{1}{2}.x+\frac{1}{4}\right)+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\) ( ddpcm )
x2 + 6x + 10
= x2 + 2 . x . 3 + 9 + 1
= (x + 3)2 + 1
(x + 3)2 lớn hơn hoặc bằng 0
(x + 3)2 + 1 lớn hơn hoặc bằng 1 > 0 (đpcm)
x2 - x + 1
= x2 - 2 . x . 1/2 + 1/4 + 3/4
= (x - 1/2)2 + 3/4
(x - 1/2)2 lớn hơn hoặc bằng 0
(x - 1/2)2 + 3/4 lớn hơn hoặc bằng 3/4 > 0 (đpcm)
a/ \(=\left(x^2+2\times x\times3+3^2\right)+1\)
\(=\left(x+3\right)^2+1\)
Ta thấy \(\left(x+2\right)^2\ge0\)
Do đó \(\left(x+3\right)^2+1\ge1>0\)
Vậy \(x^2+6x+10\) luôn luôn dương
a) \(x^2+6x+10\)
\(=\left(x^2+2\times3x+3^2\right)+1\)
\(=\left(x+3\right)^2+1\)
có (x+3)2\(\ge\)0
=>(x+3)2+1>0
=> x2+6x+10 luôn dương
b) \(x^2-x+1\)
\(=\left(x^2-2x\times\frac{1}{2}+\left(\frac{1}{2}\right)^2\right)+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)
có \(\left(x-\frac{1}{2}\right)^2\ge0\)
=> \(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)
=> x2-x+1 luôn dương
