thì x ở ngay trên số 2 mà
\(a,2^x+2^{x+3}=144\)
\(2^x+2^x\cdot2^3=144\)
\(2^x\cdot\left(1+8\right)=144\)
\(2^x\cdot9=144\)
\(2^x=144:9=16\)
\(2^x=2^4\)
\(x=16\)
\(b,2\cdot\left(3x-\frac{1}{2}\right)^2=32\)
\(\left(3x-\frac{1}{2}\right)^2=16\)
\(\left(3x-\frac{1}{2}\right)^2=\pm4^2\)
\(3x-\frac{1}{2}=\pm4\)
\(\orbr{\begin{cases}3x-\frac{1}{2}=4\\3x-\frac{1}{2}=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{7}{6}\end{cases}}}\)
Vậy...........
