Question

已知有理数a.b满足条件a^2+b^2+a^2b^2=4ab-1,求a,b的值

Answer 1

a^2+b^2+a^2b^2=4ab-1
a^2+b^2+a^2b^2-4ab+1=0
(a^2+b^2-2ab)+(a^2b^2-2ab+1)=0
(a-b)^2+(ab-1)^2=0
即
a-b=0
ab-1=0
得
a=b=1或a=b=-1

Answer 2

a^2+b^2+a^2b^2=4ab-1
a^2+b^2+a^2b^2-4ab+1=0
a^2-2ab+b^2+a^2b^2-2ab+1=0
(a-b)^2+(ab-1)^2=0
(a-b)^2=0,(ab-1)^2=0
a=b
ab-1=0
a^2-1=0
a=±1
a=b=±1
当a=1时.b=1
当a=-1时.b=-1

综上当a=1，b=1或a=-1，b=-1