\(3x^3-3x^2-3x-5=0\) (1)
Đặt \(t=x-\dfrac{1}{3}\Rightarrow x=\dfrac{1}{3}+t\) , ta được:
\(\left(1\right)\Leftrightarrow3\left(\dfrac{1}{3}+t\right)^3-3\left(\dfrac{1}{3}+t\right)^2-3\left(\dfrac{1}{3}+t\right)-5=0\)\(\Leftrightarrow3t^3-4t-\dfrac{56}{9}=0\) (2)
Đặt \(y=\dfrac{t}{\dfrac{4\sqrt{3}}{3}}\Rightarrow t=\dfrac{4\sqrt{3}}{3}y\)
\(\Rightarrow\left(2\right)\Leftrightarrow3\left(\dfrac{4\sqrt{3}}{3}y\right)^3-4\left(\dfrac{4\sqrt{3}}{3}y\right)^2-\dfrac{56}{9}=0\)\(\Leftrightarrow4y^3-3y^2=\dfrac{7\sqrt{3}}{6}\)
Đặt \(a=\sqrt[3]{\dfrac{7\sqrt{3}}{6}+\sqrt{\dfrac{7\sqrt{3}}{6}^2+1}}\) và \(\alpha=\dfrac{1}{2}\left(a-\dfrac{1}{a}\right)\) , ta được:
\(4\alpha^3-3\alpha=\dfrac{7\sqrt{3}}{6}\)
Vậy \(\alpha=y\) là nghiệm của pt
\(\Rightarrow y=\left(\sqrt[3]{\dfrac{7\sqrt{3}}{6}+\sqrt{\dfrac{7\sqrt{3}}{6}^2+1}}\right)\left(\sqrt[3]{\dfrac{7\sqrt{3}}{6}-\sqrt{\dfrac{7\sqrt{3}}{6}^2+1}}\right)\)\(=0,5034424461\)
\(\Rightarrow t=\dfrac{4\sqrt{3}}{3}y=1,162650527\)
\(\Rightarrow x=\dfrac{1}{3}+t=1,49598386\)
3x3-3x2-3x-5=0
x -3x -5=0
x-3x=5
-2x=5
x=\(\dfrac{-5}{2}\)
