1.已知x,y满足2x+y=6,x-3y=1,求7y(x-3y)²-2(3y-x)³的值2.已知x+y=2,xy=-½,求x(x+y)²(1-y)-x(y+x)²的值

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/16 15:49:09

1.已知x,y满足2x+y=6,x-3y=1,求7y(x-3y)²-2(3y-x)³的值2.已知x+y=2,xy=-½,求x(x+y)²(1-y)-x(y+x)²的值
1.已知x,y满足2x+y=6,x-3y=1,求7y(x-3y)²-2(3y-x)³的值
2.已知x+y=2,xy=-½,求x(x+y)²(1-y)-x(y+x)²的值

1.已知x,y满足2x+y=6,x-3y=1,求7y(x-3y)²-2(3y-x)³的值2.已知x+y=2,xy=-½,求x(x+y)²(1-y)-x(y+x)²的值
1
因为x-3y=1
所以7y(x-3y)^2-2(3y-x)^3=7Y+2
因为x-3y=1
所以x=3y+1
所以7y+2=2x+y
因为2x+y=6
所以原式=6
2 x(x+y)²(1-y)-x(y+x)²
=(x+y)^2(x-xy-x)
=(x+y)^2*(-xy)
=2^2*(1/2)
=2

7y(x-3y)²-2(3y-x)³
=(x-3y)²[7y-2(3y-x)]
=(x-3y)²(2x+y)
=6
x(x+y)²(1-y)-x(y+x)²
=(x+y)²[x(1-y)-x]
=(x+y)²(-xy)
=2

1、原式=7y+2
联立两式得7y=2,于是原式=4
2、原式=4x(1-y)-4x=4x(1-y-1)=-4xy=2