a)\(5x^5=x^3\Leftrightarrow5x^5-x^3=0\Leftrightarrow x^3\left(5x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{\sqrt{5}}\\x=-\dfrac{1}{\sqrt{5}}\end{matrix}\right.\)
Vì \(x\in N\Rightarrow x=0\)
b)\(\left(2x-3\right)^3=64\Leftrightarrow\left(2x-3\right)^3=4^3\)
\(\Rightarrow2x-3=4\Rightarrow x=\dfrac{7}{2}\) loại vì \(x\in N\)
c)\(\left(5x-1\right)^2=2023^0\cdot7+3^2\)
\(\Rightarrow\left(5x-1\right)^2=16\Leftrightarrow\left(5x-1\right)^2=4^2\) hoặc \(\left(5x-1\right)^2=\left(-4\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{3}{5}\left(l\right)\end{matrix}\right.\)
a) \(5x^5=x^3\)
\(x^3\left(5x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{\sqrt{5}}\left(l\right)\\x=-\dfrac{1}{\sqrt{5}}\left(l\right)\end{matrix}\right.\)
b) \(\left(2x-3\right)^3=64\)
\(2x-3=4\)
\(x=\dfrac{7}{2}\left(l\right)\)
c) \(\left(5x-1\right)^2=2023^0.7+3^2=16\)
\(\left[{}\begin{matrix}5x-1=4\\5x-1=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{3}{5}\left(l\right)\end{matrix}\right.\)
