(1)分解因式:x²-5x+4
=(x-1)(x-4)
(2)已知a²-2a+b²+2b+2=0,求a-b的值.
a²-2a+b²+2b+2=0
得(a-1)²+(b+1)²=0
所以a-1=0,b+1=0
解得a=1,b=-1
a-b=1-(-1）=2
(3)试说明:不管a,b取任何值,2a²+5b²-2ab+2a-4b+5的值始终为正数.
2a²+5b²-2ab+2a-4b+5
=a²+b²-2ab+a²+2a+1+4b²-4b+1+3
=(a-b)²+(a+1)²+(2b-1)²+3≥3
所以不管a,b取任何值,2a²+5b²-2ab+2a-4b+5的值始终为正数.