\(a,ĐK:x\le-2;x\ge2\\ PT\Leftrightarrow\sqrt{x+2}\left(\sqrt{x-2}+\sqrt{x+2}\right)=0\\ \Leftrightarrow\sqrt{x+2}=0\left(\sqrt{x-2}+\sqrt{x+2}>0\right)\\ \Leftrightarrow x=-2\left(tm\right)\)
a,ĐKXĐ:\(\left\{{}\begin{matrix}x^2-4\ge0\\x^2+4x+4\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2\ge4\\\left(x+2\right)^2\ge0\left(luôn.đúng\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)
\(\sqrt{x^2-4}+\sqrt{x^2+4x+4}=0 \)
Vì \(\sqrt{x^2-4}\ge0,\sqrt{x^2+4x+4}\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2-4}=0\\\sqrt{x^2+4x+4}=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)\left(x+2\right)=0\\\left(x+2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\x=-2\end{matrix}\right.\)
\(\Leftrightarrow x=-2\left(tm\right)\)
Vậy x=-2
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