Question

[Ngô Thành Chung]

Giải phương trình 

\(\sqrt[3]{x+1}+\sqrt[3]{3x-1}=\sqrt[3]{x-1}\)

Answer 1

[Nguyễn Ngọc Lộc]

Hệ \(\Leftrightarrow x+1+3x-1+3\sqrt[3]{\left(x+1\right)\left(3x-1\right)}\left(\sqrt[3]{x+1}+\sqrt[3]{3x-1}\right)=x-1\)

\(\Leftrightarrow3x+1+3\sqrt[3]{\left(x+1\right)\left(3x-1\right)\left(x-1\right)}=0\)

\(\Leftrightarrow3x+1=-3\sqrt[3]{\left(x+1\right)\left(3x-1\right)\left(x-1\right)}\)

\(\Leftrightarrow27x^3+9x+27x^2+1=-27\left(x^2-1\right)\left(3x-1\right)\)

\(\Leftrightarrow27x^3+9x+27x^2+1+81x^3-81x-27x^2+27=0\)

\(\Leftrightarrow108x^3-72x+28=0\)

\(\Leftrightarrow x^3-\dfrac{2}{3}x+\dfrac{7}{27}=0\)

- AD công thức các đa nô :

\(\Rightarrow x=\sqrt[3]{-\dfrac{-\dfrac{2}{3}}{2}+\sqrt{\dfrac{\left(-\dfrac{2}{3}\right)^2}{4}+\dfrac{\left(\dfrac{7}{27}\right)^3}{27}}}+\sqrt[3]{-\dfrac{-\dfrac{2}{3}}{2}-\sqrt{\dfrac{\left(-\dfrac{2}{3}\right)^2}{4}+\dfrac{\left(\dfrac{7}{27}\right)^3}{27}}}\)

\(\Rightarrow x\approx-0,96685\)