\(\left(5x-2\right)^{20}=\left(5x-2\right)^{16}\)
\(\Leftrightarrow\left(5x-2\right)^{20}-\left(5x-2\right)^{16}=0\)
\(\Leftrightarrow\left(5x-2\right)^{16}\left[\left(5x-2\right)^4-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(5x-2\right)^{16}=0\\\left(5x-2\right)^4-1=0\Rightarrow\left(5x-2\right)^4=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-2=0\\\left[{}\begin{matrix}5x-2=1\\5x-2=-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=\dfrac{1}{5}\end{matrix}\right.\end{matrix}\right.\)
Vậy..................
\(\Leftrightarrow\left(5x-2\right)^{20}-\left(5x-2\right)^{16}=0\)
\(\Leftrightarrow\left(5x-2\right)^{16}=\left(\left(5x-2\right)^{20-16}-1\right)=0\)
\(\Leftrightarrow\left(5x-2\right)^{16}=0\Rightarrow5x-2=0\Rightarrow x_1=\dfrac{2}{5}\)
Hoặc
\(\Leftrightarrow\left(\left(5x-2\right)^4-1\right)=0\Rightarrow\left|5x-2\right|=1\)
\(\Rightarrow5x-2=1\Rightarrow x_2=\dfrac{3}{5}\)
\(\Rightarrow5x-2=-1\Rightarrow x_3=\dfrac{1}{5}\)
KL:
\(x_1=\dfrac{2}{5}\\
x_2=\dfrac{3}{5}\\
x_3=\dfrac{1}{5}\)
Giải:
\(\left(5x-2\right)^{20}=\left(5x-2\right)^{16}\)
\(\Leftrightarrow\left(5x-2\right)^{20}:\left(5x-2\right)^{16}=\left(5x-2\right)^{16}:\left(5x-2\right)^{16}\)
\(\Leftrightarrow\left(5x-2\right)^4=1\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-2=1\\5x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=3\\5x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy \(x=\dfrac{3}{5}\) hoặc \(x=\dfrac{1}{5}\).
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