Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2,\\
a,\\
3x + \sqrt {3x - 7} = 7\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge \dfrac{7}{3}} \right)\\
\Leftrightarrow 3x - 7 - \sqrt {3x - 7} = 0\\
\Leftrightarrow {\sqrt {3x - 7} ^2} - \sqrt {3x - 7} = 0\\
\Leftrightarrow \sqrt {3x - 7} .\left( {\sqrt {3x - 7} - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {3x - 7} = 0\\
\sqrt {3x - 7} - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt {3x - 7} = 0\\
\sqrt {3x - 7} = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
3x - 7 = 0\\
3x - 7 = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{3}\\
x = \dfrac{8}{3}
\end{array} \right.\\
b,\\
\sqrt {4x} + \sqrt {\dfrac{x}{4}} + \dfrac{1}{2}\sqrt {49x} = 6\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 0} \right)\\
\Leftrightarrow \sqrt {{2^2}.x} + \sqrt {{{\left( {\dfrac{1}{2}} \right)}^2}.x} + \dfrac{1}{2}.\sqrt {{7^2}.x} = 6\\
\Leftrightarrow 2.\sqrt x + \dfrac{1}{2}.\sqrt x + \dfrac{1}{2}.7.\sqrt x = 6\\
\Leftrightarrow \sqrt x .\left( {2 + \dfrac{1}{2} + \dfrac{7}{2}} \right) = 6\\
\Leftrightarrow 6\sqrt x = 6\\
\Leftrightarrow \sqrt x = 1\\
\Leftrightarrow x = 1\\
3,\\
a,\\
DKXD:\,\,\,a > 0\\
Q = \left( {\dfrac{1}{{\sqrt a + 1}} - \dfrac{1}{{a + \sqrt a }}} \right):\dfrac{1}{{2 - \sqrt 5 }}\\
= \left( {\dfrac{1}{{\sqrt a + 1}} + \dfrac{1}{{\sqrt a .\left( {\sqrt a + 1} \right)}}} \right).\left( {2 - \sqrt 5 } \right)\\
= \dfrac{{\sqrt a + 1}}{{\sqrt a .\left( {\sqrt a + 1} \right)}}.\left( {2 - \sqrt 5 } \right)\\
= \dfrac{{2 - \sqrt 5 }}{{\sqrt a }}\\
b,\\
2 = \sqrt 4 < \sqrt 5 \Rightarrow 2 - \sqrt 5 < 0\\
\Rightarrow \dfrac{{2 - \sqrt 5 }}{{\sqrt a }} < 0,\,\,\,\forall a > 0\\
\Rightarrow Q < 0 \Rightarrow Q < 1
\end{array}\)
