Giải thích các bước giải:
\(\begin{array}{l}
a,\\
A = 3 + \sqrt {x - 2} \,\,\,\,\,\,\,\,\,\,\left( {x \ge 2} \right)\\
x \ge 2 \Rightarrow \sqrt {x - 2} \ge 0\\
\Rightarrow A = 3 + \sqrt {x - 2} \ge 3,\,\,\,\forall x \ge 2\\
\Rightarrow {A_{\min }} = 3 \Leftrightarrow x = 2\\
b,\\
B = \sqrt {{x^2} - 4x + 7} - 1\\
= \sqrt {\left( {{x^2} - 4x + 4} \right) + 3} - 1\\
= \sqrt {{{\left( {x - 2} \right)}^2} + 3} - 1\\
\ge \sqrt {0 + 3} - 1 = \sqrt 3 - 1\\
\Rightarrow {B_{\min }} = \sqrt 3 - 1 \Leftrightarrow {\left( {x - 2} \right)^2} = 0 \Leftrightarrow x = 2\\
c,\\
C = x - \sqrt x + 1\\
= \left( {x - \sqrt x + \dfrac{1}{4}} \right) + \dfrac{3}{4}\\
= {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge 0 + \dfrac{3}{4} = \dfrac{3}{4},\,\,\,\,\forall x \ge 0\\
\Rightarrow {C_{\min }} = \dfrac{3}{4} \Leftrightarrow {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} = 0 \Leftrightarrow x = \dfrac{1}{4}
\end{array}\)
