Question

Chứng minh:

A (x)= ^x2-4x+5>0

^{B(x)= x2}+x+1>0

C(x)= 8x - ^x2-17<0

Answer 1

* Ta có: \(A\left(x\right)=x^2-4x+5=\left(x^2-2\cdot x\cdot2+2^2\right)-2^2+5=\left(x-2\right)^2+1\ge1>0\)

Vậy \(A\left(x\right)=x^2-4x+5>0\)

b. \(B\left(x\right)=x^2+x+1=\left[x^2+2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]-\left(\dfrac{1}{2}\right)^2+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

Vậy \(B\left(x\right)=x^2+x+1>0\)

c. \(C\left(x\right)=8x-x^2-17=-x^2+8x-17=-\left(x^2-8x\right)-17=-\left(x^2-2\cdot x\cdot4+4^2\right)+4^2-17=-\left(x-4\right)^2-1\le-1< 0\)

Vậy \(C\left(x\right)=8x-x^2-17< 0\)