Đáp án:
\(\begin{array}{l}
MinA = 3\\
MinB = - 30\\
MinC = 3\\
MinD = - \dfrac{9}{4}\\
MinE = - 1001\\
MinF = - \dfrac{{41}}{{16}}\\
MinG = 2019\\
MinH = \dfrac{{16159}}{8}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = {x^2} + 2.x.3 + 9 + 3\\
= {\left( {x + 3} \right)^2} + 3\\
Do:{\left( {x + 3} \right)^2} \ge 0\forall x\\
\to {\left( {x + 3} \right)^2} + 3 \ge 3\\
\to MinA = 3\\
\Leftrightarrow x = - 3\\
B = {x^2} - 2.x.5 + 25 - 30\\
= {\left( {x - 5} \right)^2} - 30\\
Do:{\left( {x - 5} \right)^2} \ge 0\forall x\\
\to {\left( {x - 5} \right)^2} - 30 \ge - 30\\
\to MinB = - 30\\
\Leftrightarrow x = 5\\
C = 4{x^2} + 2.2x.2 + 4 + 3\\
= {\left( {2x + 2} \right)^2} + 3\\
Do:{\left( {2x + 2} \right)^2} \ge 0\forall x\\
\to {\left( {2x + 2} \right)^2} + 3 \ge 3\\
\to MinC = 3\\
\Leftrightarrow x = - 1\\
D = {x^2} - 2.x.\dfrac{5}{2} + \dfrac{{25}}{4} - \dfrac{9}{4}\\
= {\left( {x - \dfrac{5}{2}} \right)^2} - \dfrac{9}{4}\\
Do:{\left( {x - \dfrac{5}{2}} \right)^2} \ge 0\forall x\\
\to {\left( {x - \dfrac{5}{2}} \right)^2} - \dfrac{9}{4} \ge - \dfrac{9}{4}\\
\to MinD = - \dfrac{9}{4}\\
\Leftrightarrow x = \dfrac{5}{2}\\
E = 9{x^2} - 2.3x.1 + 1 - 1001\\
= {\left( {3x - 1} \right)^2} - 1001\\
Do:{\left( {3x - 1} \right)^2} \ge 0\forall x\\
\to {\left( {3x - 1} \right)^2} - 1001 \ge - 1001\\
\to MinE = - 1001\\
\Leftrightarrow x = \dfrac{1}{3}\\
F = 4{x^2} - 2.2x.\dfrac{3}{4} + \dfrac{9}{{16}} - \dfrac{{41}}{{16}}\\
= {\left( {2x - \dfrac{3}{4}} \right)^2} - \dfrac{{41}}{{16}}\\
Do:{\left( {2x - \dfrac{3}{4}} \right)^2} \ge 0\forall x\\
\to {\left( {2x - \dfrac{3}{4}} \right)^2} - \dfrac{{41}}{{16}} \ge - \dfrac{{41}}{{16}}\\
\to MinF = - \dfrac{{41}}{{16}}\\
\Leftrightarrow x = \dfrac{3}{8}\\
G = 2{x^2} + 2.x\sqrt 2 .\sqrt 2 + 2 + 2019\\
= {\left( {x\sqrt 2 + \sqrt 2 } \right)^2} + 2019\\
Do:{\left( {x\sqrt 2 + \sqrt 2 } \right)^2} \ge 0\\
\to {\left( {x\sqrt 2 + \sqrt 2 } \right)^2} + 2019 \ge 2019\\
\to MinG = 2019\\
\Leftrightarrow x = - 1\\
H = 2{x^2} + 2.x\sqrt 2 .\dfrac{3}{{2\sqrt 2 }} + \dfrac{9}{8} + \dfrac{{16159}}{8}\\
= {\left( {x\sqrt 2 + \dfrac{3}{{2\sqrt 2 }}} \right)^2} + \dfrac{{16159}}{8}\\
Do:{\left( {x\sqrt 2 + \dfrac{3}{{2\sqrt 2 }}} \right)^2} \ge 0\forall x\\
\to {\left( {x\sqrt 2 + \dfrac{3}{{2\sqrt 2 }}} \right)^2} + \dfrac{{16159}}{8} \ge \dfrac{{16159}}{8}\\
\to MinH = \dfrac{{16159}}{8}\\
\Leftrightarrow x\sqrt 2 + \dfrac{3}{{2\sqrt 2 }} = 0\\
\to x = - \dfrac{3}{4}
\end{array}\)
