\(\Leftrightarrow\left(\sqrt[3]{162x^3+2}-2\right)-\left(\sqrt{27x^2-9x+1}-1\right)=0\)
\(\Leftrightarrow\frac{\sqrt[3]{162x^3-6}}{\sqrt[3]{\left(162x^3+2\right)^2}+2\sqrt[3]{162x^3+2}+4}-\frac{27x^2-9x}{\sqrt{27x^2-9x+1}+1}=0\)
\(\Leftrightarrow\left(3x-1\right)\left[\frac{6\left(9x^2+3x+1\right)}{\sqrt[3]{\left(162x^3+2\right)^2}+2\sqrt[3]{162x^3+2}+4}-\frac{9x}{\sqrt{27x^2-9x+1}+1}\right]=0\)
\(\Leftrightarrow x=\frac{1}{3}\).
\(\left(\sqrt[3]{162x^3+2}-2\right)+\left(1-\sqrt{27x^2-9x+1}\right)=0\)
\(\frac{6\left(27x^3-1\right)}{\left(\sqrt[3]{\left(162x^3+2\right)^2}+2\sqrt[3]{162x^3}+2\right)}+\frac{-9\left(3x-1\right)}{\sqrt{27x^2-9x+1}+1}=0\)
x = 1/3
Mình chỉ làm dc đến đấy thôi.
