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

*) Tìm GTNN

\({\left( {\sqrt {x - 1} + \sqrt {3 - 2y} } \right)^2}\)\( = x - 2y + 2 + 2\sqrt {\left( {x - 1} \right)\left( {3 - 2y} \right)} \)

Ta có: \(x - 2y + 2 = 2\sqrt {x - 2y + 2 + 2\sqrt {\left( {x - 1} \right)\left( {3 - 2y} \right)} } \ge 2\sqrt {x - 2y + 2} \ge 0\)

\( \Rightarrow S = x - 2y \ge - 2\)

Dấu “\( = \)” xảy ra khi và chỉ khi \(x = 1\); \(y = \dfrac{3}{2}\)

\( \Rightarrow {\mathop{\rm m}\nolimits} = \min S = - 2\)\( \Leftrightarrow \left\{ \begin{array}{l}x = 1\\y = \dfrac{3}{2}\end{array} \right.\)

*) Tìm GTLN

Ta có:

\(\begin{array}{l}x - 2y + 2 = 2\left( {\sqrt {x - 1} + \sqrt {3 - 2y} } \right)\\ \Leftrightarrow {\left( {x - 2y + 2} \right)^2} = 4{\left( {\sqrt {x - 1} + \sqrt {3 - 2y} } \right)^2}\\ \Leftrightarrow {\left( {x - 2y + 2} \right)^2} = 4{\left( {\sqrt {x - 1} + \sqrt {3 - 2y} } \right)^2}\\ \Leftrightarrow {\left( {x - 2y + 2} \right)^2} = 4\left( {x - 1 + 3 - 2y + 2\sqrt {x - 1} .\sqrt {3 - 2y} } \right)\\ \Leftrightarrow {\left( {x - 2y + 2} \right)^2} = 4\left( {x - 2y + 2 + 2\sqrt {x - 1} .\sqrt {3 - 2y} } \right) \le 4\left( {x - 2y + 2 + x - 2y + 2} \right)\\ \Leftrightarrow {\left( {x - 2y + 2} \right)^2} \le 8\left( {x - 2y + 2} \right)\\ \Leftrightarrow x - 2y + 2 \le 8\\ \Leftrightarrow x - 2y \le 6\\ \Leftrightarrow S \le 6\end{array}\)

Dấu “\( = \)” xảy ra \( \Leftrightarrow \left\{ \begin{array}{l}x - 1 = 3 - 2y\\x - 2y = 6\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x + 2y = 4\\x - 2y = 6\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x = 5\\y = - \dfrac{1}{2}\end{array} \right.\)

\( \Rightarrow M = \max \,S = 6\)\( \Leftrightarrow \left\{ \begin{array}{l}x = 5\\y = - \dfrac{1}{2}\end{array} \right.\)

Vậy \(M + m = 6 + \left( { - 2} \right) = 4\).