已知x=根号2+1,求代数式x⼀x-1+(x-2⼀x²-1)⼀(x²-2x-2⼀x²+2x+1)的值

解：∵x=√2+1
∴x²-2x-2=(√2+1)²-2(√2+1)-2=3+2√2-2√2-4=-1
原式=x/(x-1)+(x-2)/(x+1)(x-1)×(x+1)²/(x²-2x-2)
=x/(x-1)+(x-2)(x+1)/(x-1)(x²-2x-2)
=(√2+1)/√2-(√2-1)(√2+2)/√2
=(√2+1-2-2√2+√2+2)/√2
=1/√2
=√2/2
希望采纳。

x/x-1+(x-2/x²-1)/(x²-2x-2/x²+2x+1)
=x/(x-1)+[(x-2)/(x-1)(x+1)]/[(x²-2x-2)(x+1)²]
=x/(x-1)+(x-2)(x+1)/[(x-1)(x²-2x-2)]
=[x(x²-2x-2)+(x-2)(x+1)]/(x-1)(x²-2x-2)
=[(√2+1)(3+2√2-2√2-2-2)+(√2+1-2)(√2+1+1)]/[(√2+1-1)(3+2√2-2√2-2-2)]
=[(√2+1)(-1)+(√2-1)(√2+2)]/(-√2)
=(-√2+2+√2-2)/(-√2)
=0