Đáp án:
\(\begin{array}{l}
1)21\\
2)B = {x^2} - x - 3\\
3)Max = - \dfrac{{23}}{8}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)Thay:x = - 2\\
\to A = 3.{\left( { - 2} \right)^2} - 4.\left( { - 2} \right) + 1\\
= 3.4 + 8 + 1 = 21\\
2)B = \left( {x - 1} \right)\left( {{x^2} + x + 1} \right) - \left( {x + 1} \right)\left( {2 - x} \right) - {x^3}\\
= {x^3} - 1 + {x^2} - x - 2 - {x^3}\\
= {x^2} - x - 3\\
3)C = B - A = {x^2} - x - 3 - 3{x^2} + 4x - 1\\
= - 2{x^2} + 3x - 4\\
= - \left( {2{x^2} - 3x + 4} \right)\\
= - \left( {2{x^2} - 2.x\sqrt 2 .\dfrac{3}{{2\sqrt 2 }} + \dfrac{9}{8} + \dfrac{{23}}{8}} \right)\\
= - {\left( {x\sqrt 2 - \dfrac{3}{{2\sqrt 2 }}} \right)^2} - \dfrac{{23}}{8}\\
Do:{\left( {x\sqrt 2 - \dfrac{3}{{2\sqrt 2 }}} \right)^2} \ge 0\forall x\\
\to - {\left( {x\sqrt 2 - \dfrac{3}{{2\sqrt 2 }}} \right)^2} \le 0\\
\to - {\left( {x\sqrt 2 - \dfrac{3}{{2\sqrt 2 }}} \right)^2} - \dfrac{{23}}{8} \le - \dfrac{{23}}{8}\\
\to Max = - \dfrac{{23}}{8}\\
\Leftrightarrow x\sqrt 2 - \dfrac{3}{{2\sqrt 2 }} = 0\\
\to x = \dfrac{3}{4}
\end{array}\)
