作业帮已知x=根号2+1,求代数式x/x-1+(x-2/x²-1)/(x²-2x-2/x²+2x
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已知x=根号2+1,求代数式x/x-1+(x-2/x²-1)/(x²-2x-2/x²+2x+1)的值
要详细的解法
要详细的解法
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
=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
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