\(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
\(4x^{30}-1^{30}=4x^{20}-1^{20}\)
\(4x^{30}-4x^{20}=-1+1\)
\(4x^{20}\left(x^{10}-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4x^{20}=0\\x^{10}-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x^{20}=0\\x^{10}=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}}\)
hok tốt!!
Ta có \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
<=> \(\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)
<=> \(\left(4x-1\right)^{20}.\left[\left(4x-1\right)^{10}-1\right]=0\)
<=> \(\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left(4x-1\right)^{10}-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x=1\\4x-1=1;4x-1=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\4x=2;4x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=2;x=0\end{cases}}\)
Vậy \(x\in\left\{0;2;\frac{1}{4}\right\}\)
T k chắc bài t nhưng t chắc bạn ღTĭểυ Tɦưღ lm sai ròi )) lũy thừa thì lm j cs cái ct đó
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