x^4+(1/x)^4=[x^2+(1/x)^2]^2-2*x^2*(1/x)^2
=[x^2+(1/x)^2]^2-2=[(x+1/x)^2-2*x*(1/x)]^2-2=[(x+1/x)^2-2]^2-2
现在就要求x+1/x的值,用已知x^2-4x+1=0来变形,当x=0时,原式不成立,除以x,得x-4+1/x=0则x+1/x=4,把该式代入[(x+1/x)^2-2]^2-2,得x的四次方+1/x的四次方=194
这些题,多做即可
x^4+(1/x)^4=[x^2+(1/x)^2]^2-2*x^2*(1/x)^2
=[x^2+(1/x)^2]^2-2=[(x+1/x)^2-2*x*(1/x)]^2-2=[(x+1/x)^2-2]^2-2
现在就要求x+1/x的值,用已知x^2-4x+1=0来变形,当x=0时,原式不成立,除以x,得x-4+1/x=0则x+1/x=4,把该式代入[(x+1/x)^2-2]^2-2,得x的四次方+1/x的四次方=194
这些题,多做即可