已知x=1/根号2-1,求x²-2x-1的值

已知x=1/根号2-1,求x²-2x-1的值
已知x=1/根号2-1,求x²-2x-1的值

已知x=1/根号2-1,求x²-2x-1的值
x=1/根号2-1
=(√2+1)/[(√2+1)(√2-1)]
=√2+1
x²-2x-1
=x²-2x+1-2
=(x-1)²-2
=(√2+1-1)²-2
=2-2
=0

分母有理化
x=√2+1
x-1=√2
两边平方
x²-2x+1=2
两边减去2
所以x²-2x-1=0

解
x=1/(√2-1)=(√2+1)/(√2-1)(√2+1)=1+√2
∴x²-2x-1
=(x²-2x+1)-2
=(x-1)²-2
=(1+√2-1)²-2
=(√2)²-2
=2-2
=0