Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
M = \sqrt {{x^2} - 2x + 10} = \sqrt {\left( {{x^2} - 2x + 1} \right) + 9} = \sqrt {{{\left( {x - 1} \right)}^2} + 9} \\
{\left( {x - 1} \right)^2} \ge 0,\,\,\,\forall x\\
\Rightarrow {\left( {x - 1} \right)^2} + 9 \ge 9,\,\,\,\forall x\\
\Rightarrow M = \sqrt {{{\left( {x - 1} \right)}^2} + 9} \ge \sqrt 9 = 3,\,\,\,\forall x\\
\Rightarrow {M_{\min }} = 3 \Leftrightarrow {\left( {x - 1} \right)^2} = 0 \Leftrightarrow x = 1\\
2,\\
x - 2\sqrt x - 3 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 0} \right)\\
\Leftrightarrow \left( {x - 3\sqrt x } \right) + \left( {\sqrt x - 3} \right) = 0\\
\Leftrightarrow \sqrt x \left( {\sqrt x - 3} \right) + \left( {\sqrt x - 3} \right) = 0\\
\Leftrightarrow \left( {\sqrt x - 3} \right)\left( {\sqrt x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x - 3 = 0\\
\sqrt x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \sqrt x = 3\,\,\,\,\,\,\,\,\,\,\left( {\sqrt x \ge 0} \right)\\
\Leftrightarrow x = 9\\
3,\\
\dfrac{{\sqrt {{a^3}} }}{{\sqrt a }} = \sqrt {\dfrac{{{a^3}}}{a}} = \sqrt {{a^2}} = a\,\,\,\,\,\,\,\,\left( {a > 0} \right)\\
4,\\
A = \sqrt { - {x^2} + 2x + 15} = \sqrt {16 - \left( {{x^2} - 2x + 1} \right)} = \sqrt {16 - {{\left( {x - 1} \right)}^2}} \\
{\left( {x - 1} \right)^2} \ge 0,\,\,\,\forall x\\
\Rightarrow 16 - {\left( {x - 1} \right)^2} \le 16,\,\,\,\forall x\\
\Rightarrow A = \sqrt {16 - {{\left( {x - 1} \right)}^2}} \le \sqrt {16} = 4,\,\,\,\forall x\\
\Rightarrow {A_{\max }} = 4 \Leftrightarrow {\left( {x - 1} \right)^2} = 0 \Leftrightarrow x = 1
\end{array}\)
