giúp mik bài 3b với ạ!!!! cần gấp ạ :<

Đáp án:

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