Toán Lớp 9: 3 ( căn x) / (căn ) – 2 > ( căn x) + 2
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Toán Lớp 9: 3 ( căn x) / (căn ) – 2 > ( căn x) + 2
Comments ( 2 )
\dfrac{{3\sqrt x }}{{\sqrt x – 2}} > \sqrt x + 2\\
\to \dfrac{{3\sqrt x – \left( {\sqrt x + 2} \right)\left( {\sqrt x – 2} \right)}}{{\sqrt x – 2}} > 0\\
\to \dfrac{{3\sqrt x – x + 4}}{{\sqrt x – 2}} > 0\\
\to \dfrac{{\left( {4 – \sqrt x } \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x – 2}} > 0\\
\to \dfrac{{4 – \sqrt x }}{{\sqrt x – 2}} > 0\left( {do:\sqrt x + 1 > 0} \right)\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
4 – \sqrt x > 0\\
\sqrt x – 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
4 – \sqrt x < 0\\
\sqrt x – 2 < 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
4 > \sqrt x > 2\\
\left\{ \begin{array}{l}
4 < \sqrt x \\
\sqrt x < 2
\end{array} \right.\left( l \right)
\end{array} \right.\\
\to 16 > x > 4
\end{array}\)