a: Ta có: \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{8}\\x=-\dfrac{3}{8}\end{matrix}\right.\)
b: Ta có: \(\left(\dfrac{1}{2}\right)^x=\dfrac{1}{32}\)
nên x=5
c: Ta có: \(5^{x+1}-5^x=20\)
\(\Leftrightarrow5^x=5\)
hay x=1
c: Ta có: \(2^x+2^{x+4}=544\)
\(\Leftrightarrow2^x\cdot17=544\)
\(\Leftrightarrow2^x=32\)
hay x=5
c: Ta có: \(4^{2x+1}+4^{2x}=80\)
\(\Leftrightarrow4^{2x}=16\)
\(\Leftrightarrow2x=2\)
hay x=1
c: Ta có: \(3^{2x+2}+3^{2x+1}=108\)
\(\Leftrightarrow3^{2x}=9\)
\(\Leftrightarrow2x=2\)
hay x=1
c: Ta có: \(7^{x+3}-7^{x+1}=16464\)
\(\Leftrightarrow7^x=49\)
hay x=2
