Có a^2+b^2=4. Tìm min M=2a-b

\[\begin{array}{l} {a^2} + {b^2} = 4\,\,\,\left( * \right)\\ M = 2a - b\\ \Leftrightarrow b = 2a - M\\ \Rightarrow \left( * \right) \Leftrightarrow {a^2} + {\left( {2a - M} \right)^2} = 4\\ \Leftrightarrow {a^2} + 4{a^2} - 4Ma + {M^2} = 4\\ \Leftrightarrow 5{a^2} - 4Ma + {M^2} - 4\, = 0\,\,\,\,\,\,\left( 1 \right)\\ \Rightarrow \left( 1 \right)\,\,\,\,co\,\,\,nghiem\,\,\,\, \Leftrightarrow \Delta ' \ge 0\\ \Leftrightarrow 4{M^2} - 5\left( {{M^2} - 4} \right) \ge 0\\ \Leftrightarrow 4{M^2} - 5{M^2} + 20 \ge 0\\ \Leftrightarrow {M^2} \le 20\\ \Leftrightarrow - 2\sqrt 5 \le M \le 2\sqrt 5 .\\ \Rightarrow Min\,\,M = - 2\sqrt 5 . \end{array}\]