若根号2x-y+ly+1l=0求【(x-y)的平方+(x+y)(x-y)】÷2x的平方根
问题描述:
若根号2x-y+ly+1l=0求【(x-y)的平方+(x+y)(x-y)】÷2x的平方根
答
根号2x-y+ly+1l=0
∴2x-y=0
y+1=0
∴x=-1/2
y=-1
【(x-y)的平方+(x+y)(x-y)】÷2x的平方根
=(x²-2xy+y²+x²-y²)÷2|x|
=(x²-xy)/|x|
=(1/4+1/2)/|-1/2|
=3/2对不起哈,打错了,在帮个忙呗,谢谢,我提高悬赏···· 若根号2x-y+ly+2l=0求【(x-y)的平方+(x+y)(x-y)】÷2x的平方根根号2x-y+ly+1l=0∴2x-y=0 y+2=0∴x=-1 y=-2 【(x-y)的平方+(x+y)(x-y)】÷2x的平方根=(x²-2xy+y²+x²-y²)÷2|x|=(x²-xy)/|x|=(1-2)/|-1|=-1