k) ĐXKĐ: \(x+2\ne0\Leftrightarrow x\ne-2\)
\(\dfrac{x+2}{3}=\dfrac{3}{x+2}\)
\(\Leftrightarrow\left(x+2\right)\left(x+2\right)=3.3\\ \Leftrightarrow\left(x+2\right)^2=3^2\\ \Leftrightarrow\left[{}\begin{matrix}x+2=-3\\x+2=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-5\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{-5,1\right\}\)
l) ĐXKĐ: \(x-4\ne0\Leftrightarrow x\ne4\)
\(\dfrac{x-4}{-5}=\dfrac{-5}{x-4}\Leftrightarrow\left(x-4\right)\left(x-4\right)=\left(-5\right).\left(-5\right)\\ \Leftrightarrow\left(x-4\right)^2=5^2\\ \Leftrightarrow\left[{}\begin{matrix}x-4=-5\\x-4=5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=9\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{-1,9\right\}\)
k)\(\dfrac{x+2}{3}=\dfrac{3}{x+2}\)
\(\Leftrightarrow\left(x+2\right)^2=9\)
\(\Rightarrow\left[{}\begin{matrix}x+2=3\\x+2=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
Vậy \(x\in\left\{1;-5\right\}\)
l)\(\dfrac{x-4}{-5}=\dfrac{-5}{x-4}\)
\(\Leftrightarrow\left(x-4\right)^2=25\)
\(\Rightarrow\left[{}\begin{matrix}x-4=5\\x-4=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=9\\x=-1\end{matrix}\right.\)
Vậy \(x\in\left\{9;-1\right\}\)
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