a) \(-\left(x^2-6x+10\right)=-\left(x^2-6x+9+1\right)=-\left[\left(x-3\right)^2+1\right]\le-1< 0\forall x\)
BĐT đúng
b) \(x^2+x+1=x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)
BĐT đúng
c)Dấu "=" ko xảy ra???
\(=\left(4x^2+2.2x.y+y^2\right)+2\left(2x+y\right)+1+2\)
\(=\left(2x+y\right)^2+2.\left(2x+y\right).1+1+1\)
\(=\left(2x+y+1\right)^2+1\ge1>0\) (đpcm)
a. −x2 + 6x - 10
= −(x2 − 6x) − 10
= −(x2 − 2.x.3 + 32 − 9) − 10
= −(x − 3)2 + 9 − 10
= −(x − 3)2 −1
Vì (x − 3)2 ≥ 0 ∀ x ⇒ −(x − 3)2 ≤ 0 ⇒ −(x − 3)2 −1 ≤ −1
Vậy −(x − 3)2 −1 < 0 ⇒ −x2 + 6x - 10 luôn âm với mọi x
b. x2 + x + 1
= x2 + 2.x.\(\frac{1}{2}\)+ (\(\frac{1}{2}\))2 − \(\frac{1}{4}\) + 1
= (x + \(\frac{1}{2}\))2 + \(\frac{3}{4}\)
Vì (x + \(\frac{1}{2}\))2 ≥ 0 ∀ x ⇒ (x + \(\frac{1}{2}\))2 + \(\frac{3}{4}\) ≥ \(\frac{3}{4}\) ∀ x
Vậy (x + \(\frac{1}{2}\))2 + \(\frac{3}{4}\) ≥ 0 hay x2 + x + 1 > 0 ∀ x.
Câu c hơi khó 1 chút nhé xin lỗi nhé :v
