\(\frac{x+1}{16}=\frac{4}{x+1}\\ \Rightarrow\left(x+1\right)^2=64\\ \Rightarrow\left[{}\begin{matrix}x+1=8\\x+1=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-9\end{matrix}\right.\)
Vậy \(x\in\left\{7;-9\right\}\)
\(\frac{x+1}{16}=\frac{4}{x+1}\)
\(\Rightarrow\left(x+1\right)^2=4.16\)
\(\Rightarrow\left(x+1\right)^2=64\)
\(\Rightarrow\left(x+1\right)^2=8^2\)
\(\Rightarrow\left[{}\begin{matrix}x+1=8\\x+1=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-9\end{matrix}\right.\)
Vậy \(x=7\) hoặc \(x=-9\)
\(\frac{x+1}{16}=\frac{4}{x+1}\)
\(\Rightarrow\left(x+1\right).\left(x+1\right)=4.16\)
\(\Rightarrow\left(x+1\right).\left(x+1\right)=64\)
\(\Rightarrow\left(x+1\right)^2=64\)
\(\Rightarrow\left(x+1\right)^2=\left(\pm8\right)^2\)
\(\Rightarrow x+1=\pm8.\)
\(\Rightarrow\left[{}\begin{matrix}x+1=8\\x+1=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8-1\\x=\left(-8\right)-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-9\end{matrix}\right.\)
Vậy \(x\in\left\{7;-9\right\}.\)
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