\(\begin{array}{l}
\text{Câu 2:}\\
a)\quad \sqrt{(2x+3)^2} = 2018\\
\Leftrightarrow |2x + 3| = 2018\\
\Leftrightarrow \left[\begin{array}{l}2x + 3 =2018\\2x + 3 =-2018\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \dfrac{2015}{2}\\x = -\dfrac{2021}{2}\end{array}\right.\\
\text{Vậy}\ S = \left\{-\dfrac{2021}{2};\dfrac{2015}{2}\right\}\\
b)\quad \sqrt{9x} -5\sqrt x = 6 - 4\sqrt x\quad (ĐK: x \geqslant 0)\\
\Leftrightarrow 3\sqrt x - 5\sqrt x + 4\sqrt x = 6\\
\Leftrightarrow \sqrt x = 3\\
\Rightarrow x = 9\quad (nhận)\\
\text{Vậy}\ S = \{9\}\\
\text{Câu 3:}\\
\quad Q = \left(\dfrac{1}{\sqrt a + 1} - \dfrac{1}{a + \sqrt a}\right):\dfrac{\sqrt a - 1}{a + 2\sqrt a + 1}\\
a)\quad \text{Điều kiện:}\ \begin{cases}a \geqslant 0\\a + \sqrt a \ne 0\end{cases}\Leftrightarrow a >0\\
\quad Q = \left[\dfrac{1}{\sqrt a + 1} - \dfrac{1}{\sqrt a\left(\sqrt a + 1\right)}\right]:\dfrac{\sqrt a - 1}{\left(\sqrt a + 1\right)^2}\\
\Leftrightarrow Q = \dfrac{\sqrt a - 1}{\sqrt a\left(\sqrt a + 1\right)}\cdot \dfrac{\left(\sqrt a + 1\right)^2}{\sqrt a - 1}\\
\Leftrightarrow Q = \dfrac{\sqrt a +1}{\sqrt a}\\
\Leftrightarrow Q = 1 + \dfrac{1}{\sqrt a}\\
b)\quad \text{Xét}\ Q - 1\\
= 1 + \dfrac{1}{\sqrt a} - 1\\
= \dfrac{1}{\sqrt a}\\
\text{Do}\ a >0\\
\text{nên}\ \sqrt a >0\\
\Leftrightarrow \dfrac{1}{\sqrt a}>0\\
\text{hay}\ Q - 1 >0\\
\text{Vậy}\ Q >1\\
c)\quad \text{Ta có:}\\
a = - 7\sqrt[3]{49\left(5 + 4\sqrt2\right)\left(3 + 2\sqrt{1 + \sqrt2}\right)\left(3 -2\sqrt{1 + \sqrt2}\right)}\\
\Leftrightarrow a = - 7\sqrt[3]{49\left(5 + 4\sqrt2\right)\left[9 -4\left(1 + \sqrt2\right)\right]}\\
\Leftrightarrow a = - 7\sqrt[3]{49\left(5 + 4\sqrt2\right)\left(5 - 4\sqrt2\right)}\\
\Leftrightarrow a = - 7\sqrt[3]{49\left(25 - 16.2\right)}\\
\Leftrightarrow a = - 7\sqrt[3]{49.(-7)}\\
\Leftrightarrow a = - 7.(-7)\\
\Leftrightarrow a = 49\\
\Rightarrow \sqrt a = 7\\
\text{Thay vào Q ta được:}\\
Q = \dfrac{7 + 1}{7} = \dfrac87
\end{array}\)
