化简求值(x²-4/x²-x-1)²÷(x³-3x²+2x/x³-1)²(x/x+2)³其中x=
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题目,应该 是这样的吧
(x²-4/x²-x+1)²÷(x³-3x²+2x/x³-1)²(x/x+2)³ 其中x= -2/3
原式={[(x+2)(x-2)/(x^2-x+1)]*[(x-1)(x^2+x+1)/x(x-)(x-2)]}^2*(x/x+2)³
=[(x+2)/(x-2)]^2*(x/x+2)³
=x^3/(x+2)(x-2)
当x=-2/3时,原式=1/12
(x²-4/x²-x+1)²÷(x³-3x²+2x/x³-1)²(x/x+2)³ 其中x= -2/3
原式={[(x+2)(x-2)/(x^2-x+1)]*[(x-1)(x^2+x+1)/x(x-)(x-2)]}^2*(x/x+2)³
=[(x+2)/(x-2)]^2*(x/x+2)³
=x^3/(x+2)(x-2)
当x=-2/3时,原式=1/12
追问
怎么做
追答
(x²-4/x²-x+1)²÷(x³-3x²+2x/x³-1)²(x/x+2)³ 其中x= -2/3
原式={[(x+2)(x-2)/(x^2-x+1)]*[(x-1)(x^2+x+1)/x(x-)(x-2)]}^2*(x/x+2)³
=[(x+2)/(x-2)]^2*(x/x+2)³
=x^3/[(x+2)(x-2)]
当x=-2/3时,原式=(-8/27)*(-9/32)=1/12
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(x²-4/x²-x-1)²÷(x³-3x²+2x/x³-1)²(x/x+2)³
=[(x+2)(x-2)/(x^2-x-1)]^2*[(x-1)(x^2+x+1)/x(x-2)(x-1)]^2*[x^3/(x+2)^3]
=[(x+2)^2(x-2)^2/(x^-x-1)^2]*[(x^2+x+1)^2/x^2(x-2)^2]*[x^3/(x+2)^3
=x(x^2-x-1)^2/(x+2)(x^2+x+1)^2=[x/(x+2)]*[(x^2-x-1)/(x^2+x+1)]^2
代x=-2/3
得原式=[(-2/3)/(2-2/3)]*[(1/9)/(19/9)]^2
=(-1/4)*(1/19)^2=1/1444
=[(x+2)(x-2)/(x^2-x-1)]^2*[(x-1)(x^2+x+1)/x(x-2)(x-1)]^2*[x^3/(x+2)^3]
=[(x+2)^2(x-2)^2/(x^-x-1)^2]*[(x^2+x+1)^2/x^2(x-2)^2]*[x^3/(x+2)^3
=x(x^2-x-1)^2/(x+2)(x^2+x+1)^2=[x/(x+2)]*[(x^2-x-1)/(x^2+x+1)]^2
代x=-2/3
得原式=[(-2/3)/(2-2/3)]*[(1/9)/(19/9)]^2
=(-1/4)*(1/19)^2=1/1444
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题写的额部清楚,特别是除号所针对的分母分子,不清楚
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