Question

chứng minh rằng 

A= x²-x+1>0,∀x

B= x²+x+1>0,∀x

C= x²+2x+2>0,∀x

Answer 1

`A=x^2 -x +1 = x^2 -2*x*1/2 +(1/2)^2 +1 -(1/2)^2`

`A=(x-1/2)^2 + 3/4 >0 AAx `

Vậy `A=x^2 -x +1 >0`

`B=x^2 +x+1 = x^2 +2*x*1/2 +(1/2)^2 +1 -(1/2)^2`

`B= (x+1/2)^2 +3/4 > 0 AAx`

Vậy `B=x^2 +x +1 >0`

`C=x^2 +2x+2 = x^2 +2*x*1 +1^2 +2-1^2`

`C=(x+1)^2 +1 >0`

Vậy `C=x^2 +2x+2>0`

Answer 2

a: \(=x^2-x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

b: \(=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

c: \(=x^2+2x+1+1=\left(x+1\right)^2+1>0\)