Đáp án:
\(Vay\,\,\mathop {Max}\limits_{\left[ {2;\,\,4} \right]} A = 2\,\,\,\,khi\,\,\,x = 3.\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \sqrt {x - 2} + \sqrt {4 - x} \\
DK:\,\,\,2 \le x \le 4\\
\Rightarrow {A^2} = x - 2 + 4 - x + 2\sqrt {\left( {x - 2} \right)\left( {4 - x} \right)} \\
\Leftrightarrow {A^2} = 2 + 2\sqrt { - \left( {{x^2} - 6x + 8} \right)} \\
Ta\,\,co:\,\,\,{x^2} - 6x + 8 = {x^2} - 6x + 9 - 1 = {\left( {x - 3} \right)^2} - 1\\
\Rightarrow - \left( {{x^2} - 6x + 8} \right) = - {\left( {x - 3} \right)^2} + 1 \le 1\\
\Rightarrow \sqrt { - \left( {{x^2} - 6x + 8} \right)} \le 1\\
\Rightarrow 2\sqrt { - \left( {{x^2} - 6x + 8} \right)} \le 2\\
\Rightarrow 2 + 2\sqrt { - \left( {{x^2} - 6x + 8} \right)} \le 4\\
\Rightarrow {A^2} \le 4.\\
Dau\,\, = \,\,\,xay\,\,ra \Leftrightarrow x - 3 = 0 \Leftrightarrow x = 3\,\,\,\left( {tm} \right)\\
Vay\,\,\mathop {Max}\limits_{\left[ {2;\,\,4} \right]} A = 2\,\,\,\,khi\,\,\,x = 3.
\end{array}\)
