Question

Giải phương trình :

\(x^3+2x^2-4x=-\dfrac{8}{3}\)

Answer 1

 

.

Answer 2

`x^3 + 2x^2 - 4x + 8/3 = 0`

`<=>` \(\left(x+\dfrac{2\left(\sqrt[3]{2}+1\right)^2}{3}\right)^3=0\)

`<=>`\(x=-\dfrac{2\left(\sqrt[3]{2}+1\right)^2}{3}^3\)

Answer 3

\(\Leftrightarrow3x^3+6x^2-12x+8=0\)

\(\Leftrightarrow4x^3-\left(x^3-6x^2+12x-8\right)=0\)

\(\Leftrightarrow4x^3-\left(x-2\right)^3=0\)

\(\Leftrightarrow\left(\sqrt[3]{4}x\right)^3=\left(x-2\right)^3\)

\(\Leftrightarrow\sqrt[3]{4}x=x-2\)

\(\Leftrightarrow x=\dfrac{-2}{\sqrt[3]{4}-1}\)