\(a,2^{x-1}=16\\ 2^{x-1}=2^4\\ x-1=4\\ x=4+1\\ x=5\\ b,\left|x+3\right|=7\\ TH1:x+3=7\\ x=7-3\\ x=4\\ TH2:x+3=-7\\ x=-7-3\\ x=-10\\ c,\dfrac{1}{5}+x=\dfrac{1}{2}\\ x=\dfrac{1}{2}-\dfrac{1}{5}\\ x=\dfrac{3}{10}\)
\(2^{x-1}=16\)
\(\Rightarrow2^{x-1}=2^4\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=5\)
Vậy \(x=5\)
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\(\left|x+3\right|=7\)
\(\Rightarrow\left[{}\begin{matrix}x+3=7\\x+3=-7\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=4\\x=-10\end{matrix}\right.\)
Vậy \(x\in\left\{4;-10\right\}\)
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\(\dfrac{1}{5}+x=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}-\dfrac{1}{5}\)
\(x=\dfrac{3}{10}\)
Vậy \(x=\dfrac{3}{10}\)
