Đáp án:
$\begin{array}{l}
b)Dkxd:\left\{ \begin{array}{l}
4x + 1 \ge 0\\
3x - 2 \ge 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ge - \dfrac{1}{4}\\
x \ge \dfrac{2}{3}
\end{array} \right. \Rightarrow x \ge \dfrac{2}{3}\\
\sqrt {4x + 1} - \sqrt {3x - 2} = \dfrac{{x + 3}}{5}\\
\Leftrightarrow \dfrac{{4x + 1 - \left( {3x - 2} \right)}}{{\sqrt {4x + 1} + \sqrt {3x - 2} }} - \dfrac{{x + 3}}{5} = 0\\
\Leftrightarrow \dfrac{{x + 3}}{{\sqrt {4x + 1} + \sqrt {3x - 2} }} - \dfrac{{x + 3}}{5} = 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {\dfrac{1}{{\sqrt {4x + 1} + \sqrt {3x - 2} }} - \dfrac{1}{5}} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x + 3 = 0\\
\sqrt {4x + 1} + \sqrt {3x - 2} = 5
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - 3\left( {ktm} \right)\\
4x + 1 + 2\sqrt {4x + 1} .\sqrt {3x - 2} + 3x - 2 = 25
\end{array} \right.\\
\Rightarrow 7x - 1 + 2\sqrt {12{x^2} - 5x - 2} = 25\\
\Rightarrow 2\sqrt {12{x^2} + 5x - 2} = 26 - 7x\\
\left( {dk:\dfrac{2}{3} \le x \le \dfrac{{26}}{7}} \right)\\
\Rightarrow 4\left( {12{x^2} + 5x - 2} \right) = {\left( {26 - 7x} \right)^2}\\
\Rightarrow 48{x^2} + 20x - 8 = 49{x^2} - 364x + 676\\
\Rightarrow {x^2} - 384x + 684 = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 1,789\left( {tm} \right)\\
x = 382,21\left( {ktm} \right)
\end{array} \right.\\
Vay\,x = 1,789
\end{array}$
