Question

已知x=(根号2+1)的-1次方,y=(根号2-1)的-1次方,求(x+1)的-2次方+(y-1)的-2次方的值

Answer 1

因为1/x=√2+1
所以x=√2-1
同理y=√2+1
所以xy=2-1=1
x-y=2
平方
x²+y²-2xy=4
x²+y²=4+2xy=6
原式=1/(x+1)²+1/(y-1)²
=(x²+2x+1+y²-2y+1)/(xy+x-y-1)²
=[(x²+y²)+2(x-y)+2]/[xy+(x-y)-1]²
=12/2²
=3
你看我打字这么累你给我吧

Answer 2

1/x=√2+1
所以x=√2-1
同理y=√2+1
所以xy=2-1=1
x-y=2
平方
x²+y²-2xy=4
x²+y²=4+2xy=6

原式=1/(x+1)²+1/(y-1)²
=(x²+2x+1+y²-2y+1)/(xy+x-y-1)²
=[(x²+y²)+2(x-y)+2]/[xy+(x-y)-1]²
=12/2²
=3

Answer 3

x=(√2+1)^(-1)=1/(√2+1)=√2-1 //分式上下同乘√2-1
y=(√2-1)^(-1)=1/(√2-1)=√2+1 //分式上下同乘√2+1

(x+1)^(-2)+(y-1)^(-2)
=1/(x+1)²+1/(y-1)²
=1/2+1/2
=1