Tìm $n$ để: $D[(n^2-n)^3]=3$

Đáp án:

`D[(n^2-n)^3]=3`

`<=> 1/[log_2[(n^2-n)^3])=3`
`<=> log[(n^2-n)^3]/log2=1/3`
`<=> log[(n^2-n)^3]=log2/3`
`<=> log[(n^2-n)^3]=log(\root(3)(2))`
`<=> (n^2-n)^3=\root(3)(2)`
`<=>`  \(\left[ \begin{array}{l}
n^2 - n = \sqrt[9]{2}
\\
n^2 - n = \sqrt[9]{2}\cdot e^{\frac{2\pi i}{3}}
\\
n^2 - n = \sqrt[9]{2}\cdot e^{\frac{4\pi i}{3}}
\end{array} \right.\) 
`<=>`\(\left[ \begin{array}{l}
n^2 - n+\frac{1}{4} = \sqrt[9]{2}+\frac{1}{4}
\\
n^2 - n+\frac{1}{4} = \sqrt[9]{2}\cdot e^{\frac{2\pi i}{3}}+\frac{1}{4}

\\
n^2 - n+\frac{1}{4} = \sqrt[9]{2}\cdot e^{\frac{4\pi i}{3}}+\frac{1}{4}

\end{array} \right.\) 
`<=>`\(\left[ \begin{array}{l}
(n-\frac{1}{2})^2 = \sqrt[9]{2}+\frac{1}{4}
\\
(n-\frac{1}{2})^2 = \sqrt[9]{2}\cdot e^{\frac{2\pi i}{3}}+\frac{1}{4}

\\
(n-\frac{1}{2})^2 = \sqrt[9]{2}\cdot e^{\frac{4\pi i}{3}}+\frac{1}{4}

\end{array} \right.\) 
`<=>`\(\left[ \begin{array}{l}
n-\frac{1}{2} = \sqrt{\sqrt[9]{2}+\frac{1}{4}}
\\
n-\frac{1}{2} = -\sqrt{\sqrt[9]{2}+\frac{1}{4}}
\\
n-\frac{1}{2} = \sqrt{\sqrt[9]{2}\cdot e^{\frac{2\pi i}{3}}+\frac{1}{4}}
\\
n-\frac{1}{2} = -\sqrt{\sqrt[9]{2}\cdot e^{\frac{2\pi i}{3}}+\frac{1}{4}}
\\
n-\frac{1}{2} = \sqrt{\sqrt[9]{2}\cdot e^{\frac{4\pi i}{3}}+\frac{1}{4}}
\\
n-\frac{1}{2} = -\sqrt{\sqrt[9]{2}\cdot e^{\frac{4\pi i}{3}}+\frac{1}{4}}
\end{array} \right.\) 
`<=>`\(\left[ \begin{array}{l}
n= \sqrt{\sqrt[9]{2}+\frac{1}{4}}+\frac{1}{2} 
\\
n = -\sqrt{\sqrt[9]{2}+\frac{1}{4}}+\frac{1}{2}
\\
n= \sqrt{\sqrt[9]{2}\cdot e^{\frac{2\pi i}{3}}+\frac{1}{4}}+\frac{1}{2} 
\\
n= -\sqrt{\sqrt[9]{2}\cdot e^{\frac{2\pi i}{3}}+\frac{1}{4}}+\frac{1}{2} 
\\
n= \sqrt{\sqrt[9]{2}\cdot e^{\frac{4\pi i}{3}}+\frac{1}{4}}+\frac{1}{2} 
\\
n = -\sqrt{\sqrt[9]{2}\cdot e^{\frac{4\pi i}{3}}+\frac{1}{4}}+\frac{1}{2}
\end{array} \right.\)