Question

已知x=√2-1,y=√2+1,求下列各式的值。⑴（x-1）（y-1） ⑵y/x+x/y

Answer 1

(x - 1)(y - 1) = xy - x - y + 1
=(√2 - 1)(√2 + 1) - (√2 - 1) -(√2 + 1) + 1
= 1 - 2√2 + 1 = 2 - 2√2
y/x + x/y = y²/xy + x²/xy
= (y² + x²)/xy
=[ (√2+1)² + (√2-1)²]/(√2-1)(√2+1)
= 6

Answer 2

原式=xy-x-y+1
=（√2-1）（√2+1）-（√2-1）-（√2+1）+1
=1-2√2+1
=2-2√2
原式=(x^2+y^2)/(xy)
=[(x+y)^2-2xy]/xy
=(x+y)^2/xy-2
=(√2-1+√2+1)^2/(√2-1)(√2+1)-2
=8