Question

[Queen Material]

Tìm x :
3x³ - 3x² - 3x - 5= 0

Answer 1

[T.Thùy Ninh]

\(3x^3-3x^2-3x-5=0\) (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

\(\Rightarrow y=\left(\sqrt[3]{\dfrac{7\sqrt{3}}{6}+\sqrt{\dfrac{7\sqrt{3}}{6}^2+1}}\right)\left(\sqrt[3]{\dfrac{7\sqrt{3}}{6}-\sqrt{\dfrac{7\sqrt{3}}{6}^2+1}}\right)\)\(=0,5034424461\)

\(\Rightarrow t=\dfrac{4\sqrt{3}}{3}y=1,162650527\)

\(\Rightarrow x=\dfrac{1}{3}+t=1,49598386\)

Answer 2

[Thảo Nguyễn]

3x³-3x²-3x-5=0

x -3x -5=0

x-3x=5

-2x=5

x=\(\dfrac{-5}{2}\)