Đáp án:
c) x=4
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:x = 49\\
\to R = \dfrac{{2.49}}{{\sqrt {49} + 3}} = \dfrac{{49}}{5}\\
b)S = \dfrac{{\left( {1 - 3\sqrt x } \right)\left( {\sqrt x + 3} \right) + 4x + 14\sqrt x + 6}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{ - 3x - 8\sqrt x + 3 + 4x + 14\sqrt x + 6}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x + 6\sqrt x + 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{{{\left( {\sqrt x + 3} \right)}^2}}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x + 3}}{{\sqrt x - 3}}\\
c)R.S = - \sqrt x - 6\\
\to \dfrac{{2x}}{{\sqrt x + 3}}.\dfrac{{\sqrt x + 3}}{{\sqrt x - 3}} = - \sqrt x - 6\\
\to \dfrac{{2x}}{{\sqrt x - 3}} = - \sqrt x - 6\\
\to 2x = \left( {\sqrt x - 3} \right)\left( { - \sqrt x - 6} \right)\\
\to 2x = - x - 3\sqrt x + 18\\
\to 3x + 3\sqrt x - 18 = 0\\
\to 3\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right) = 0\\
\to \left[ \begin{array}{l}
\sqrt x = 2\\
\sqrt x = - 3\left( l \right)
\end{array} \right.\\
\to x = 4
\end{array}\)
