a) \(\left(2x-1\right)^{10}=\left(1-2x\right)^5\)
\(\Rightarrow\left(2x-1\right)^2=1-2x\)
\(\Rightarrow4x^2-4x+1=1-2x\)
\(\Rightarrow4x^2-4x=-2x\)
\(\Rightarrow2x^2-2x=-x\)
\(\Rightarrow2x^2-x=0\)
\(\Rightarrow x.\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
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b) \(\left(3x-1\right)^{15}=\left(1-3x\right)^8\)
\(\Rightarrow\left(3x-1\right)^{15}-\left(1-3x\right)^8=0\)
\(\Rightarrow\left(3x-1\right)^{15}-\left(-\left(3x-1\right)\right)^8=0\)
\(\Rightarrow\left(3x-1\right)^{15}-\left(3x-1\right)^8=0\)
\(\Rightarrow\left(3x-1\right)^8.\left(\left(3x-1\right)^7-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(3x-1\right)^8=0\\\left(3x-1\right)^7-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy ....
c) Tự lm
