a, \(\left(2x-1\right)^6=\left(2x-1\right)^7\)
\(\Rightarrow\left(2x-1\right)^6-\left(2x-1\right)^7=0\)
\(\Rightarrow\left(2x-1\right)^6.\left[1-\left(2x-1\right)\right]=0\)
\(\Rightarrow\hept{\begin{cases}\left(2x-1\right)^6=0\\1-\left(2x-1\right)=0\end{cases}\Rightarrow\hept{\begin{cases}x=0,5\\x=1\end{cases}\Rightarrow}x\in\left\{0,5;1\right\}}\)
\(\left(x-3\right)^{-2}=\frac{1}{9}\)
b, \(\Rightarrow\frac{1}{\left(x-3\right)^2}=\frac{1}{9}\)
\(\Rightarrow\left(x-3\right)^2=9\)
\(\Rightarrow\sqrt{\left(x-3\right)^2}=\pm\sqrt{9}\)
\(\Rightarrow\left(x-3\right)=\pm3\)
\(\Rightarrow\hept{\begin{cases}x-3=-3\\x-3=3\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=6\end{cases}\Rightarrow}x\in\left\{0;6\right\}}\)
có ai có thể giúp mình nữa ko bài lớp 7 này ko làm ơn
