Số nghiệm của phương trình \(4x\sqrt {x + 3}  + 2\sqrt {2x - 1}  = 4{x^2} + 3x + 3\) là

ĐKXĐ:\(\left\{ \begin{array}{l}x + 3 \ge 0\\2x - 1 \ge 0\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x \ge  - 3\\x \ge \dfrac{1}{2}\end{array} \right.\)\( \Leftrightarrow x \ge \dfrac{1}{2}\)

\(\begin{array}{l}4x\sqrt {x + 3}  + 2\sqrt {2x - 1}  = 4{x^2} + 3x + 3\\ \Leftrightarrow 4{x^2} - 2.2x.\sqrt {x + 3}  + {\left( {\sqrt {x + 3} } \right)^2} - x - 3 - 2\sqrt {2x - 1}  + 3x + 3 = 0\\ \Leftrightarrow 4{x^2} - 2.2x.\sqrt {x + 3}  + {\left( {\sqrt {x + 3} } \right)^2} + 2x - 2\sqrt {2x - 1}  = 0\\ \Leftrightarrow 4{x^2} - 2.2x.\sqrt {x + 3}  + {\left( {\sqrt {x + 3} } \right)^2} + 2x - 1 - 2\sqrt {2x - 1}  + 1 = 0\\ \Leftrightarrow 4{x^2} - 2.2x.\sqrt {x + 3}  + {\left( {\sqrt {x + 3} } \right)^2} + {\left( {\sqrt {2x - 1} } \right)^2} - 2.1.\sqrt {2x - 1}  + 1 = 0\\ \Leftrightarrow {\left( {2x - \sqrt {x + 3} } \right)^2} + {\left( {\sqrt {2x - 1}  - 1} \right)^2} = 0\end{array}\)

Ta có:

\(\left. \begin{array}{l}{\left( {2x - \sqrt {x + 3} } \right)^2} \ge 0\\{\left( {\sqrt {2x - 1}  - 1} \right)^2} \ge 0\end{array} \right\}\)\( \Rightarrow {\left( {2x - \sqrt {x + 3} } \right)^2} + {\left( {\sqrt {2x - 1}  - 1} \right)^2} \ge 0\)

Dấu “\( = \)” xảy ra khi và chỉ khi:

 \(\begin{array}{l}\,\left\{ \begin{array}{l}2x - \sqrt {x + 3}  = 0\\\sqrt {2x - 1}  - 1 = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{\left( {2x} \right)^2} = {\left( {\sqrt {x + 3} } \right)^2}\\{\left( {\sqrt {2x - 1} } \right)^2} = 1\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}4{x^2} - x - 3 = 0\\2x = 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x = 1\\x =  - \dfrac{3}{4}\end{array} \right.\\x = 1\end{array} \right.\\ \Leftrightarrow x = 1\,\left( {tm} \right)\end{array}\)

Vậy phương trình có 1 nghiệm là \(x = 1\).