Bài 1:
b: Ta có: \(\left(2x-1\right)^{10}=49^5\)
\(\Leftrightarrow\left(2x-1\right)^2=49\)
\(\Leftrightarrow\left(2x-8\right)\left(2x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
c: Ta có: \(3^x+3^{x+2}=\dfrac{10\left(6^3+3\cdot6^2+3^3\right)}{13}\)
\(\Leftrightarrow3^x\cdot10=\dfrac{3510}{13}=270\)
\(\Leftrightarrow3^x=27\)
hay x=3
