\(\sqrt{x+24}+\sqrt{x-16}=10\) (ĐK: \(x\ge16\))
\(\Leftrightarrow\left(\sqrt{x+24}+\sqrt{x-16}\right)^2=10^2\)
\(\Leftrightarrow x+24+x-16+2\sqrt{\left(x+24\right)\left(x-16\right)}=100\)
\(\Leftrightarrow2x+8+2\sqrt{\left(x+24\right)\left(x-16\right)}=100\)
\(\Leftrightarrow2\sqrt{\left(x+24\right)\left(x-16\right)}=100-2x-8\)
\(\Leftrightarrow2\sqrt{x^2-16x+24x-384}=92-2x\)
\(\Leftrightarrow\sqrt{x^2+8x-384}=46-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}46-x\ge0\\x^2+8x-384=\left(46-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}16\le x\le46\\x^2+8x-384=2116-92x+x^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}16\le x\le46\\8x+92x=2116+384\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}16\le x\le46\\100x=2500\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}16\le x\le46\\x=25\left(tm\right)\end{matrix}\right.\)
\(\Leftrightarrow x=25\)
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