Question

已知x=1+根号2,y=1-根号2,求(x-y分之1-x+y分之1)÷x²-2xy+y²分之2y的值

Answer 1

x=1+根号2,y=1-根号2,

∴x+y=2

x-y=2√2

(x-y分之1-x+y分之1)÷x²-2xy+y²分之2y

=[1/(x-y)-1/(x+y)×(x-y)²/2y

=(x+y-x+y)/(x+y)(x-y)×(x-y)²/2y

=2y/(x+y)×(x-y)/2y

=(x-y)/(x+y)

=2√2/2

=√2