Question

已知(a+b)的2次方=7,(a-b)的2次方=4,求ab的值

要有过程，谢谢

Answer 1

(a+b)²=a²+2ab+b²=7
(a-b)²=a²-2ab+b²=4
两式相减
(a²+2ab+b²)-(a²-2ab+b²)=7-4
4ab=3
ab=3/4

Answer 2

(a+b)的2次方=7,(a-b)的2次方=4
(a+b)的2次方-(a-b)的2次方
=4ab=7-4=3
ab=3/4

Answer 3

(a+b)^2=a^2+2ab+b^2=7
(a-b)^2=a^2-2ab+b^2=4
(a+b)^2-(a-b)^2=4ab=7-4=3
ab=3/4 对的

Answer 4

(a+b)^2 = 7
(a-b)^2 = 4
ab = [(a+b)^2-(a-b)^2]/4 = (7-4)/4 = 3/4

Answer 5

解答：
(a+b)^2=7
(a-b)^2=4
(1)a^2+2ab+b^2=7
(2)a^2-2ab+b^2=4
由(1)-(2)
2ab+2ab=7-4
4ab=3
ab=3/4
所以A*B=3/4

Answer 6

解：
A平方+2AB+B的平方=7；A平方-2AB+B平方=4
4AB=3，A*B=3/4