\(\left(x+3\right)\left(x^2-3x+9\right)-x^2=2x\)
\(\Leftrightarrow x^3-27-x^2-2x=0\)
\(\Leftrightarrow x^3-x^2-2x-27=0\)(1)
Đặt : \(t=x-\dfrac{1}{3}\Rightarrow x=t+\dfrac{1}{3}\) , ta được:
\(\left(1\right)\Leftrightarrow\left(t+\dfrac{1}{3}\right)^3-\left(t+\dfrac{1}{3}\right)^2-2\left(t+\dfrac{1}{3}\right)-27=0\)
\(\Leftrightarrow t^3-\dfrac{7}{3}t-\dfrac{749}{27}=0\) (2)
Đặt \(n=\dfrac{t}{\dfrac{2\sqrt{7}}{3}}\Rightarrow x=t.\dfrac{2\sqrt{7}}{3}\)
Khi đó :
\(\left(2\right)\Leftrightarrow\left(n.\dfrac{2\sqrt{7}}{3}\right)^2-\dfrac{7}{3}\left(n.\dfrac{2\sqrt{7}}{3}\right)-\dfrac{749}{27}=0\)\(\Leftrightarrow4n^3-3n=20,22109931\)
Đặt \(a=\sqrt[3]{20,22109931+\sqrt{20,22109931^2+1}}\)
\(\alpha=\dfrac{1}{2}\left(a-\dfrac{1}{a}\right)\) , ta được:
\(4\alpha^3+3\alpha=n\Rightarrow n=a\)
Vậy nghiệm của pt là:
\(n=\dfrac{1}{2}\left(\sqrt[3]{20,22109931+\sqrt{20,22109931^2+1}}+\sqrt[3]{20,22109931-\sqrt{20,22109931^2+1}}\right)\)\(=1,570967184\)
\(\Rightarrow\) \(t=1,570967184.\dfrac{2\sqrt{7}}{3}=2,770925657\)
\(\Rightarrow x=2,770925657+\dfrac{1}{3}=3,104258991\)
