Ta có: \(\left(2x-1\right)^{10}=\left(2x-1\right)^{11}\)
\(\Leftrightarrow\left(2x-1\right)^{10}-\left(2x-1\right)^{11}=0\)
\(\Leftrightarrow\left(2x-1\right)^{10}\cdot\left[1-\left(2x-1\right)\right]=0\)
\(\Leftrightarrow\left(2x-1\right)^{10}\cdot\left(1-2x+1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)^{10}\cdot\left(-2x+2\right)=0\)
\(\Leftrightarrow\left(2x-1\right)^{10}\cdot2\cdot\left(1-x\right)=0\)
Vì 2≠0
nên \(\left[{}\begin{matrix}\left(2x-1\right)^{10}=0\\1-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=1\end{matrix}\right.\)
Vì x∈Z nên x=1
Vậy: x=1
\(\left(2x-1\right)^{10}\) = \(\left(2x-1\right)^{11}\)
\(\Rightarrow\left(2x-1\right)^{11}-\left(2x-1\right)^{10}=0\) (vì A = B thì A - B = 0)
\(\Rightarrow\left(2x-1\right)^{11}.1-\left(2x-1\right)^{10}.\left(2x-1\right)=0\)
\(\Rightarrow\left(2x-1\right)^{11}.\left[1-\left(2x-1\right)\right]=0\)
\(\Rightarrow\left(2x-1\right)^{11}=0\) hoặc \(1-\left(2x-1\right)=0\)
\(\Rightarrow\) \(\left(2x-1\right)^{11}=0^{11}\) hoặc \(2x-1\) = 1
\(\Rightarrow\) \(\left(2x-1\right)^{11}=0^{11}\) hoặc \(2x\) = 1+1
\(\Rightarrow\) \(\left(2x-1\right)=0\) hoặc \(2x\) = 2
\(\Rightarrow2x=0+1\) hoặc \(x\) = 2:2
\(\Rightarrow2x=1\) hoặc \(x\) = 1
\(\Rightarrow\) x = 1:2 hoặc x =1
\(\Rightarrow x=0.5\) hoặc x = 1
Vì x \(\in\) Z \(\Rightarrow\) x=1
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