Đáp án:
\(\begin{array}{l}
MinA = 1\\
MinC = 2\\
MinB = \sqrt 7 \\
MinD = 2
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \sqrt {{x^2} - 6x + 9 + 1} \\
= \sqrt {{{\left( {x - 3} \right)}^2} + 1} \\
Do:{\left( {x - 3} \right)^2} \ge 0\forall x\\
\to {\left( {x - 3} \right)^2} + 1 \ge 1\\
\to \sqrt {{{\left( {x - 3} \right)}^2} + 1} \ge 1\\
\to Min = 1\\
\Leftrightarrow x = 3\\
C = \sqrt {2\left( {{x^2} + 4x + 6} \right)} \\
= \sqrt {2\left( {{x^2} + 4x + 4 + 2} \right)} \\
= \sqrt {2{{\left( {x + 2} \right)}^2} + 4} \\
Do:2{\left( {x + 2} \right)^2} \ge 0\forall x\\
\to 2{\left( {x + 2} \right)^2} + 4 \ge 4\\
\to \sqrt {2{{\left( {x + 2} \right)}^2} + 4} \ge 2\\
\to Min = 2\\
\Leftrightarrow x = - 2\\
B = \sqrt {3\left( {{x^2} - 2x} \right) + 10} \\
= \sqrt {3\left( {{x^2} - 2x + 1} \right) - 3 + 10} \\
= \sqrt {3{{\left( {x - 1} \right)}^2} + 7} \\
Do:3{\left( {x - 1} \right)^2} \ge 0\forall x\\
\to 3{\left( {x - 1} \right)^2} + 7 \ge 7\\
\to \sqrt {3{{\left( {x - 1} \right)}^2} + 7} \ge \sqrt 7 \\
\to Min = \sqrt 7 \\
\Leftrightarrow x = 1\\
D = \sqrt {{{\left( {x + 1} \right)}^2}} + \sqrt {{{\left( {x - 1} \right)}^2}} \\
= \left| {x + 1} \right| + \left| {x - 1} \right|\\
\to \left[ \begin{array}{l}
D = x + 1 + x - 1\left( {DK:x \ge 1} \right)\\
D = x + 1 - x + 1\left( {DK:x < 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
D = 2x\\
D = 2
\end{array} \right.\\
\to MinD = 2\\
\Leftrightarrow x = 1
\end{array}\)
