5a^+2b^+1=6ab+4a-2b首先移项5a^+2b^+1-6ab-4a+2b=0拆项：(4a^2+b^1+1-4ab-4a+2b)+(a^2+b^2-2ab)=0即(2a-b-1)^2 + (a-b)^2=0平方数总是非负,所以必有2a-b-1=a-b=0即 a=b,b=2a-1解得【a=b=1】...