a)vi (2x-6)^7=(2x-6)^5
=>2x-6=0 hoac 2x-6=1
=>x=3hoac x=7/2
b)(7-2x)^9=(7-2x)^3
=>7-2x=0 hoac 7-2x=1
=>x=7/2 hoac x=3
a) \(\left(2x-6\right)^7=\left(2x-6\right)^5\)
\(\Leftrightarrow\left[2\left(x-3\right)\right]^7=\left[2\left(x-3\right)\right]^5\)
\(\Leftrightarrow2^7\left(x-3\right)^7=2^5\left(x-3\right)^5\)
\(\Leftrightarrow128\left(x-3\right)^7=32\left(x-3\right)^5\)
\(\Leftrightarrow4\left(x-3\right)^7=\left(x-3\right)^5\)
\(\Leftrightarrow4\left(x-3\right)^7-\left(x-3\right)^5=0\)
\(\Leftrightarrow\left(x-3\right)^5\left[4\left(x-3\right)-1\right]=0\)
\(\Leftrightarrow\hept{\begin{cases}x-3=0\\4\left(x-3\right)-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\x=\frac{7}{2}\\x=\frac{5}{2}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=3\\x=\frac{7}{2}\\x=\frac{5}{2}\end{cases}}\)
