若有理数x.y.z满足等式【x-1】^2+【2x-y】^4+lx-3zl=0,试求【x+y】z^2的值
问题描述：
若有理数x.y.z满足等式【x-1】^2+【2x-y】^4+lx-3zl=0,试求【x+y】z^2的值
答
【x-1】^2+【2x-y】^4+lx-3zl=0
【x-1】^2=0,【2x-y】^4=0,lx-3zl=0
x=1
【2-y】^4=0,l1-3zl=0
y=2,z=1/3
【x+y】z^2
=【1+2】(1/3)^2
=3*1/9
=1/3