Đáp án:
$MinM = \dfrac{3}{4} \Leftrightarrow a = \dfrac{5}{4};b=\dfrac{3}{4}$
Giải thích các bước giải:
Ta có:
$a,b>0$
Và:
$\begin{array}{l}
{a^2} + 2ab + {b^2} = a + b + 2\\
\Leftrightarrow {\left( {a + b} \right)^2} - \left( {a + b} \right) - 2 = 0\\
\Leftrightarrow \left( {a + b - 2} \right)\left( {a + b + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
a + b - 2 = 0\left( c \right)\\
a + b + 1 = 0\left( l \right)
\end{array} \right.\\
\Leftrightarrow a + b = 2\\
\Leftrightarrow b = 2 - a
\end{array}$
Khi đó:
$\begin{array}{l}
M = {a^2} + 3{b^2} + 2a - 5\\
= {a^2} + 3{\left( {2 - a} \right)^2} + 2a - 5\\
= {a^2} + 3\left( {{a^2} - 4a + 4} \right) + 2a - 5\\
= 4{a^2} - 10a + 7\\
= {\left( {2a} \right)^2} - 2.2a.\dfrac{5}{2} + {\left( {\dfrac{5}{2}} \right)^2} + 7 - {\left( {\dfrac{5}{2}} \right)^2}\\
= {\left( {2a - \dfrac{5}{2}} \right)^2} + \dfrac{3}{4}
\end{array}$
Mà ta có:
$\begin{array}{l}
{\left( {2a - \dfrac{5}{2}} \right)^2} \ge 0,\forall a\\
\Rightarrow {\left( {2a - \dfrac{5}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4}\\
\Rightarrow M \ge \dfrac{3}{4}\\
\Rightarrow MinM = \dfrac{3}{4}
\end{array}$
Dấu bằng xảy ra
$\begin{array}{l}
\Leftrightarrow 2a - \dfrac{5}{2} = 0\\
\Leftrightarrow a = \dfrac{5}{4}
\end{array}$
$\to b=\dfrac{3}{4}$
Vậy $MinM = \dfrac{3}{4} \Leftrightarrow a = \dfrac{5}{4}; b=\dfrac{3}{4}$
