解
∵a-b=2,a-c=1
∴(a-b)-(a-c)=2-1=1
∴c-b=1
(2a-b-c)²+(c-b)²
=[(a-b)+(a-c)]²+1²
=(2+1)²+1
=9+1
=10a^2+b^2+2a-4b+5=0求2a^2+4b-3的值解a²+b²+2a-4b+5=0(a²+2a+1)+(b²-4b+4)=0(a+1)²+(b-2)²=0∴a+1=0，b-2=0∴a=-1，b=2∴2a²+4b-3=2×1+4×2-3=7