已知x/(x^2-4x-1)=-1/3,求x^4+2x+1/x^5的值
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![已知x/(x^2-4x-1)=-1/3,求x^4+2x+1/x^5的值](/uploads/image/z/6943834-10-4.jpg?t=%E5%B7%B2%E7%9F%A5x%2F%28x%5E2-4x-1%29%3D-1%2F3%2C%E6%B1%82x%5E4%2B2x%2B1%2Fx%5E5%E7%9A%84%E5%80%BC)
已知x/(x^2-4x-1)=-1/3,求x^4+2x+1/x^5的值
已知x/(x^2-4x-1)=-1/3,求x^4+2x+1/x^5的值
已知x/(x^2-4x-1)=-1/3,求x^4+2x+1/x^5的值
x/(x^2-4x-1)=-1/3
x^2-4x-1=-3x
x^2-x-1=0
x^2=x+1
x^4=x^2+2x+1=x+1+2x+1=3x+2
x^5=x^4*x=(3x+2)x=3x^2+2x=3(x+1)+2x=5x+3
(x^4+2x+1)/x^5
=(3x+2+2x+1)/(5x+3)
=(5x+3)/(5x+3)
=1
由已知变形可得x^2=x+1
所求式分子x^4+2x+1=(x+1)^2+2x+1=x^2+4x+2=(x+1)+4x+2=5x+3
分母x^5=(x+1)^2*x=(x^2+2x+1)*x=(x+1+2x+1)*x=3x^2+2x=3(x+1)+2x=5x+3
所以原式=1