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已知x平方+3x-8=0,求代数式1/x-2*x^2-4x+4/x+1-x-1/x+2的值
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已知x平方+3x-8=0,求代数式1/x-2*x^2-4x+4/x+1-x-1/x+2的值
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答案和解析
x^2+3x-8=0
x^2+3x=8
1/(x-2) * (x^2-4x+4)/(x+1) - (x-1)/(x+2)
= 1/(x-2) * (x-2)^2/(x+1) - (x-1)/(x+2)
= (x-2)/(x+1) - (x-1)/(x+2)
= { (x-2)(x+2)-(x-1)(x+1) } / {(x+2)(x+1)}
= (x^2-4-x^2+1) / (x^2+3x+2)
= -3 / (8+2)
=-3/10
x^2+3x=8
1/(x-2) * (x^2-4x+4)/(x+1) - (x-1)/(x+2)
= 1/(x-2) * (x-2)^2/(x+1) - (x-1)/(x+2)
= (x-2)/(x+1) - (x-1)/(x+2)
= { (x-2)(x+2)-(x-1)(x+1) } / {(x+2)(x+1)}
= (x^2-4-x^2+1) / (x^2+3x+2)
= -3 / (8+2)
=-3/10
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