Ta có : \(x^3-27x^2+90x-81=0\)
=> \(x^3-3.x^2.9+3.x.9^2-729+648=0\)
=> \(\left(x-9\right)\left(x^2+9x+81\right)+648=0\)
=> \(\left(x-9\right)\left(x^2+9x+81\right)=-648\)
Ta thấy : \(x^2+9x+81=x^2+\frac{2.x.9}{2}+\frac{81}{4}+\frac{243}{4}\)
\(=\left(x+\frac{9}{2}\right)^2+\frac{243}{4}\ge\frac{243}{4}>0\)
=> \(x-9< 0\)
=> \(\left(x-9\right)\left(\left(x+\frac{9}{2}\right)^2+\frac{243}{4}\right)\le\frac{243\left(x-9\right)}{4}\)
- Dấu " = " xảy ra <=> \(\frac{243\left(x-9\right)}{4}=0\)
=> \(x-9=0\) ( KTM )
Vậy phương trình vô nghiệm .
Hic hỏi ngu quá hay sao mà bạn kia xoá bài giải luôn r 
