Question

Chứng minh:

a) \(a^2-6a+10>0\)

b) \(a^2+a+1>0\)

c) \(\left(x-3\right)\left(x-5\right)+4>0\)

Answer 1

a,\(a^2-6a+10=a^2-6a+9+1=\left(a-3\right)^2+1\)

Mà \(\left(a-3\right)^2\ge0=>\left(a-3\right)^2+1>0\)

\(=>a^2-6a+10>0\)

b, \(a^2+a+1=a^2+2a\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(a+\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(a+\frac{1}{2}\right)^2\ge0=>\left(a+\frac{1}{2}\right)+\frac{3}{4}>0\)

\(=>a^2+a+1>0\)

\(\left(x-3\right)\left(x-5\right)+4=x^2-8x+15+4\)

\(=x^2+8x+16+3=\left(x+4\right)^2+3\)

Vì \(\left(x+4\right)^2\ge0=>\left(x+4\right)^2+3>0\)

\(=>\left(x-3\right)\left(x-5\right)+4>0\)

Answer 2

chứng minh:4x²-5x+13>0

Answer 3

tìm x thỏa mãn (x-2)(5-x)