\(\Leftrightarrow\left(x-1\right)^5\cdot\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x-1\right)^5\cdot\left(x-2\right)\left(x-1+1\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
\(\left(x-1\right)^5=\left(x-1\right)^7\)
\(\left(x-1\right)^7-\left(x-1\right)^5=0\)
\(\left(x-1\right)^5\left[\left(x-1\right)^2-1\right]=0\)
\(\left(x-1\right)^5=0\) hoặc \(\left(x-1\right)^2-1=0\)
*) \(\left(x-1\right)^5=0\)
\(x-1=0\)
\(x=0+1\)
\(x=1\)
*) \(\left(x-1\right)^2-1=0\)
\(\left(x-1\right)^2=0+1\)
\(\left(x-1\right)^2=1\)
\(\left(x-1\right)^2=1^2\) hoặc \(\left(x-1\right)^2=\left(-1\right)^2\)
\(x-1=1\) hoặc \(x-1=-1\)
+) \(x-1=1\)
\(x=1+1\)
\(x=2\)
+) \(x-1=-1\)
\(x=-1+1\)
\(x=0\)
Vậy \(x=0;x=1;x=2\)
