Question

.若x（y-1）-y（x-1）=4，则二分之一（x+y）的平方-2xy=？

Answer 1

解：原式=1/2*(x^2+y^2+2xy)-2xy=1/2*(x^2+y^2)-xy
由已知得，y-x=4，则有（y-x）^2=16
即 y^2+x^2-2xy=16
所以有y^2+x^2=2xy+16 。
将此代入原式得，原式=1/2*(2xy+16 )-xy=xy+8-xy=8

Answer 2

y-x=4;
0.5*(x+y)^2-2xy=0.5*(x-y)^2=8;

Answer 3

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

(x+y)²/2-2xy
=[(x²+2xy+y²-4xy]/2
=(x²-2xy+y²)/2
=(x-y)²/2
=(-4)²/2
=16/2
=8

Answer 4

由已知得出y=x+4,带入所求方程式，得出8

Answer 5

等于8

Answer 6

结果是8，过程如下：
由于x(y-1)-y(x-1)=4
xy-x-yx+y=4
y-x=4
所以y*y-2xy+x*x=16 即y*y+x*x=16+2xy

(x+y)*(x+y)/2-2xy=(x*x+2xy+y*y)/2-2xy=(x*x+y*y)/2-xy=(16+2xy)/2-xy=8+xy-xy=8