Đáp án:

1) \(Min =  - 26\)

2) \(Min = \dfrac{{67}}{4}\)

3) \(Min =  - \dfrac{1}{8}\)

4) \(Max =  - 2\)

5) \(Max = \dfrac{{81}}{8}\)

6) \(Max = \dfrac{4}{3}\)

7) \(Max = \dfrac{{60}}{{173}}\)

Giải thích các bước giải:

\(\begin{array}{l}
1)A = {x^2} + 2.x.5 + 25 - 26\\
 = {\left( {x + 5} \right)^2} - 26\\
Do:{\left( {x + 5} \right)^2} \ge 0\forall x\\
 \to {\left( {x + 5} \right)^2} - 26 \ge  - 26\\
 \to Min =  - 26\\
 \Leftrightarrow x =  - 5\\
2)B = 4{x^2} + 2.2x.\dfrac{3}{2} + \dfrac{9}{4} + \dfrac{{67}}{4}\\
 = {\left( {2x + \dfrac{3}{2}} \right)^2} + \dfrac{{67}}{4}\\
Do:{\left( {2x + \dfrac{3}{2}} \right)^2} \ge 0\\
 \to {\left( {2x + \dfrac{3}{2}} \right)^2} + \dfrac{{67}}{4} \ge \dfrac{{67}}{4}\\
 \to Min = \dfrac{{67}}{4}\\
 \Leftrightarrow x =  - \dfrac{3}{4}\\
3)C = 2{x^2} + 2.x\sqrt 2 .\dfrac{3}{{2\sqrt 2 }} + \dfrac{9}{8} - \dfrac{1}{8}\\
 = {\left( {x\sqrt 2  + \dfrac{3}{{2\sqrt 2 }}} \right)^2} - \dfrac{1}{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{1}{8} \ge  - \dfrac{1}{8}\\
 \to Min =  - \dfrac{1}{8}\\
 \Leftrightarrow x\sqrt 2  + \dfrac{3}{{2\sqrt 2 }} = 0\\
 \to x =  - \dfrac{3}{4}\\
4)D =  - \left( {{x^2} - 2x + 1} \right) - 2\\
 =  - {\left( {x - 1} \right)^2} - 2\\
Do:{\left( {x - 1} \right)^2} \ge 0\forall x\\
 \to  - {\left( {x - 1} \right)^2} \le 0\\
 \to  - {\left( {x - 1} \right)^2} - 2 \le  - 2\\
 \to Max =  - 2\\
 \Leftrightarrow x = 1\\
5)F =  - \left( {2{x^2} - 2.x\sqrt 2 .\dfrac{5}{{2\sqrt 2 }} + \dfrac{{25}}{8}} \right) + \dfrac{{81}}{8}\\
 =  - {\left( {x\sqrt 2  - \dfrac{5}{{2\sqrt 2 }}} \right)^2} + \dfrac{{81}}{8}\\
Do:{\left( {x\sqrt 2  - \dfrac{5}{{2\sqrt 2 }}} \right)^2} \ge 0\forall x\\
 \to  - {\left( {x\sqrt 2  - \dfrac{5}{{2\sqrt 2 }}} \right)^2} \le 0\\
 \to  - {\left( {x\sqrt 2  - \dfrac{5}{{2\sqrt 2 }}} \right)^2} + \dfrac{{81}}{8} \le \dfrac{{81}}{8}\\
 \to Max = \dfrac{{81}}{8}\\
 \Leftrightarrow x\sqrt 2  - \dfrac{5}{{2\sqrt 2 }} = 0\\
 \to x = \dfrac{5}{4}\\
6)Do:{\left( {x + 2} \right)^2} \ge 0\forall x\\
 \to {\left( {x + 2} \right)^2} + 3 \ge 3\\
 \to \dfrac{4}{{{{\left( {x + 2} \right)}^2} + 3}} \le \dfrac{4}{3}\\
 \to Max = \dfrac{4}{3}\\
 \Leftrightarrow x =  - 2\\
7)G = \dfrac{{15}}{{3{x^2} + 2.x\sqrt 3 .\dfrac{{3\sqrt 3 }}{2} + \dfrac{{27}}{4} + \dfrac{{173}}{4}}}\\
 = \dfrac{{15}}{{{{\left( {x\sqrt 3  + \dfrac{{3\sqrt 3 }}{2}} \right)}^2} + \dfrac{{173}}{4}}}\\
Do:{\left( {x\sqrt 3  + \dfrac{{3\sqrt 3 }}{2}} \right)^2} \ge 0\forall x\\
 \to {\left( {x\sqrt 3  + \dfrac{{3\sqrt 3 }}{2}} \right)^2} + \dfrac{{173}}{4} \ge \dfrac{{173}}{4}\\
 \to \dfrac{{15}}{{{{\left( {x\sqrt 3  + \dfrac{{3\sqrt 3 }}{2}} \right)}^2} + \dfrac{{173}}{4}}} \le \dfrac{{60}}{{173}}\\
 \to Max = \dfrac{{60}}{{173}}\\
 \Leftrightarrow x\sqrt 3  + \dfrac{{3\sqrt 3 }}{2} = 0\\
 \to x =  - \dfrac{3}{2}
\end{array}\)