Question

已知实数a,b满足（a+b)²=1,（a-b)²=5,求a²+b+ab的值?

Answer 1

题目应是：已知实数a,b满足（a+b)²=1,（a-b)²=5,a²+b²+ab的值

(a+b)^2=1

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

(a-b)^2=5

a^2-2ab+b^2=5

两式相加2a^2+2b^2=6

a^2+b^2=3

两式相减4ab=-4

ab=-1

a^2+b^2+ab=3-1=2