已知x+y分之2x-y=2,求代数式x+y分之4x-2x-y分之4x+4y的值

问题描述:

已知x+y分之2x-y=2,求代数式x+y分之4x-2x-y分之4x+4y的值

2x-y+1的绝对值+ (3x+2分之3y)平方=0
则2x-y+1=0
3x+3y/2=0,
则2x-y=-1①
2x+y=0②
①+②,得4x=-1,则x=-1/4
①-②,得-2y=-1,则y=-1/2
x+y分之y的平方除以(x-y分之x-1)(x-x-y分之x平方)
=x+y分之y的平方除以(x-y分之y)(x-y分之-xy)
=-x³/(x+y)
=(1/4)³/(-1/4-1/2)
=-1/48

x+y分之4x-2y-2x-y分之4x+4y
=2(2x-y)/(x+y)-4(x+y)/(2x-y)
=2*2-4*(1/2)
=4-2
=2