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Điều kiện $\begin{cases}x>3\\ x^2 \ge 16 \end{cases}\Leftrightarrow
\begin{cases}x>3 \\ \left[ {\begin{matrix} x\ge4\\ x\le-4
\end{matrix}} \right. \end{cases}\Leftrightarrow x>4$
BPT
$\Leftrightarrow \dfrac{2\sqrt{x^{2}-16}}{\sqrt{x-3}}+\dfrac{x-3}{\sqrt{x-3}} > \dfrac{7-x}{\sqrt{x-3}}$
$\Leftrightarrow 2\sqrt{x^{2}-16}+x-3 > 7-x$
$\Leftrightarrow 2\sqrt{x^{2}-16} > 10-2x$
$\Leftrightarrow \sqrt{x^{2}-16} > 5-x$
$\Leftrightarrow \left[\begin{array}{l}5-x\le0\\\begin{cases}5-x>0\\x^{2}-16>(5-x)^2\end{cases}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x\ge5\\\begin{cases}x<5\\x^2-16>x^2-10x+25\end{cases}\end{array}\right.$
$\Leftrightarrow
\left[\begin{array}{l}x\ge5\\\begin{cases}x<5\\x>\dfrac{41}{10}\end{cases}\end{array}\right.
\Leftrightarrow x>\dfrac{41}{10}$
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