Question

根据条件,求下列代数式的值:(1)若x(y-1)-y(x-1)=4,求x的平方+y的平方/2-xy的值；

Answer 1

(1)若x(y-1)-y(x-1)=4,求(x²+y²)/2 - xy的值

x(y-1)-y(x-1)=4

(xy-x) - (xy-y) = 4

得到 y-x=4

(x²+y²)/2 - xy

=(x²+y²- 2xy)/2

=(y-x)² / 2

=16/2

=8

（2）若a+b=5,ab=3,求代式a³b-2a² b² +ab³

a³b-2a² b² +ab³=ab(a² -2ab +b²)

=ab[(a+b)²-4ab]

= 3×(25-4×3)

=39