Question

chứng minh biểu thức luôn dương

a) A = x^2-3x+8

b) b = 2x^2-2x+2

mình cần kết quả 1 cách nhanh nhất

 

Answer 1

a) \(x^2-3x+8=\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{23}{4}=\left(x-\dfrac{3}{2}\right)^2+\dfrac{23}{4}\ge\dfrac{23}{4}>0\)

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

Answer 2

a: Ta có: \(A=x^2-3x+8\)

\(=x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{23}{4}\)

\(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{23}{4}>0\forall x\)

b: Ta có: \(B=2x^2-2x+2\)

\(=2\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\right)\)

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