若x=√2，求代数式[(x^2-2x+4)/(√x^2-4x+4)]/[(x^3+8)/(x^2-4)]*[|6-x|/(x^2-5x-6)-[1/(x^2-x+1)]^-1的值
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解：[(x^2-2x+4)/(√x^2-4x+4)]/[(x^3+8)/(x^2-4)]*[|6-x|/(x^2-5x-6)- [1/(x^2-x+1)]^-1
=[(x^2-2x+4)/|x-2|]/[(x^3+8)/(x^2-4)]*[|6-x|/(x-6)(x+1)-(x^2- x+1)
=(x^2-2x+4)/(x-2)×(x-2)(x+2)/(x^3+8)[(6-x)/(x-6)(x+1)-(x^2- x+1)
=-(x^2-2x+4)(x+2)/(x^3+8)/(x+1)-(x^2- x+1)
=-1/(x+1)-(x^2- x+1)
x=√2
-1/(x+1)-(x^2- x+1)=-1/(√2+1)-(2-√2+1)=1-√2-3+√2=-2
=[(x^2-2x+4)/|x-2|]/[(x^3+8)/(x^2-4)]*[|6-x|/(x-6)(x+1)-(x^2- x+1)
=(x^2-2x+4)/(x-2)×(x-2)(x+2)/(x^3+8)[(6-x)/(x-6)(x+1)-(x^2- x+1)
=-(x^2-2x+4)(x+2)/(x^3+8)/(x+1)-(x^2- x+1)
=-1/(x+1)-(x^2- x+1)
x=√2
-1/(x+1)-(x^2- x+1)=-1/(√2+1)-(2-√2+1)=1-√2-3+√2=-2
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