giúp ạ hứa cho 5 sao

Điều kiện xác định $x\ge \dfrac{2}{3}$

$\begin{array}{l}
d)\sqrt {2x - 2 + 2\sqrt {2x - 3} }  + \sqrt {2x + 13 + 8\sqrt {2x - 3} }  = 5\\
 \Leftrightarrow \sqrt {\left( {2x - 3} \right) + 2\sqrt {2x - 3}  + 1}  + \sqrt {2x - 3 + 2.4\sqrt {2x - 3}  + 16}  = 5\\
 \Leftrightarrow \sqrt {{{\left( {\sqrt {2x - 3}  + 1} \right)}^2}}  + \sqrt {{{\left( {\sqrt {2x - 3}  + 4} \right)}^2}}  = 5\\
 \Leftrightarrow \left| {\sqrt {2x - 3}  + 1} \right| + \left| {\sqrt {2x - 3}  + 4} \right| = 5\\
 \Leftrightarrow \sqrt {2x - 3}  + 1 + \sqrt {2x - 3}  + 4 = 5\\
 \Leftrightarrow 2\sqrt {2x - 3}  = 0\\
 \Leftrightarrow 2x - 3 = 0 \Leftrightarrow x = \dfrac{3}{2}
\end{array}$