Question

已知a²b²+a²+b²+1=4ab,求a、b的值.

Answer 1

a^2b^2+a^2+b^2+1=4ab

a^2b^2+a^2+b^2+1=2ab+2ab

a^2b^2+a^2+b^2+1-2ab-2ab=0

(a^2b^2-2ab+1)+(a^2+b^2-2ab)=0

(ab-1)^2+(a-b)^2=0

(ab-1)^2=0

得到方程组:

ab=1 (1)

a=b (2)

(2)代入(1)得:

a^2=1

a^2-1=0

(a-1)(a+1)=0

a1=1,a2=-1

解方程组得:

a1=1,a2=-1

b1=1,b2=-1