先化再求:(1/x-y-1/x+y)÷2y/x的平方+2xy+y的平方,其中x=√3+√2,y=√3-√2

问题描述:

先化再求:(1/x-y-1/x+y)÷2y/x的平方+2xy+y的平方,其中x=√3+√2,y=√3-√2

(1/x-y-1/x+y)÷2y/x²+2xy+y²
=0÷2y/x²+2xy+y²
=2xy+y²
=2(√3+√2)(√3-√2)+(√3-√2)²
=2+3-2√6+2
=7-2√6