\(2^{x+1}+5.2^{x-2}=\dfrac{7}{32}\) ( đề sai )
Ta có:\(2^{x+1}+5.2^{x-1}=\dfrac{7}{32}\)
\(2^{x+1}.\left(1+5\right)=\dfrac{7}{32}\)
\(2^{x-1}=\dfrac{7}{32}:6\)
\(2^{x-1}=\dfrac{7}{192}\)
mà \(2^{x-1}\ne\dfrac{7}{192}\)
\(\Rightarrow x\in\varnothing\)