Đáp án:
\(\begin{array}{l}
B14:\\
c)Min = - \dfrac{{19}}{2}\\
d)Min = - 1\\
e)Min = - 11\\
m)Min = - \dfrac{9}{8}\\
B15:\\
a)Max = 2\\
c)Max = 9
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B14:\\
c)C = 2{x^2} + 6x - 5\\
= 2{x^2} + 2.x\sqrt 2 .\dfrac{3}{{\sqrt 2 }} + \dfrac{9}{2} - \dfrac{{19}}{2}\\
= {\left( {x\sqrt 2 + \dfrac{3}{{\sqrt 2 }}} \right)^2} - \dfrac{{19}}{2}\\
Do:{\left( {x\sqrt 2 + \dfrac{3}{{\sqrt 2 }}} \right)^2} \ge 0\forall x\\
\to {\left( {x\sqrt 2 + \dfrac{3}{{\sqrt 2 }}} \right)^2} - \dfrac{{19}}{2} \ge - \dfrac{{19}}{2}\\
\to Min = - \dfrac{{19}}{2}\\
\Leftrightarrow x\sqrt 2 + \dfrac{3}{{\sqrt 2 }} = 0\\
\to x = - \dfrac{3}{2}\\
d)D = 4{x^2} - 4x\\
= 4{x^2} - 4x + 1 - 1\\
= {\left( {2x - 1} \right)^2} - 1\\
Do:{\left( {2x - 1} \right)^2} \ge 0\forall x\\
\to {\left( {2x - 1} \right)^2} - 1 \ge - 1\\
\to Min = - 1\\
\Leftrightarrow x = \dfrac{1}{2}\\
e)E = {x^2} - 8x + 5\\
= \left( {{x^2} - 8x + 16} \right) - 11\\
= {\left( {x - 4} \right)^2} - 11\\
Do:{\left( {x - 4} \right)^2} \ge 0\forall x\\
\to {\left( {x - 4} \right)^2} - 11 \ge - 11\\
\to Min = - 11\\
\Leftrightarrow x = 4\\
m)M = 2{x^2} + x - 1\\
= 2{x^2} + 2.x\sqrt 2 .\dfrac{1}{{2\sqrt 2 }} + \dfrac{1}{8} - \dfrac{9}{8}\\
= {\left( {x\sqrt 2 + \dfrac{1}{{2\sqrt 2 }}} \right)^2} - \dfrac{9}{8}\\
Do:{\left( {x\sqrt 2 + \dfrac{1}{{2\sqrt 2 }}} \right)^2} \ge 0\forall x\\
\to {\left( {x\sqrt 2 + \dfrac{1}{{2\sqrt 2 }}} \right)^2} - \dfrac{9}{8} \ge - \dfrac{9}{8}\\
\to Min = - \dfrac{9}{8}\\
\Leftrightarrow x\sqrt 2 + \dfrac{1}{{2\sqrt 2 }} = 0\\
\to x = - \dfrac{1}{4}\\
B15:\\
a)A = - {x^2} - 4x - 2\\
= - \left( {{x^2} + 4x + 2} \right)\\
= - \left( {{x^2} + 4x + 4 - 2} \right)\\
= - {\left( {x + 2} \right)^2} + 2\\
Do:{\left( {x + 2} \right)^2} \ge 0\forall x\\
\to - {\left( {x + 2} \right)^2} \le 0\\
\to - {\left( {x + 2} \right)^2} + 2 \le 2\\
\to Max = 2\\
\Leftrightarrow x = - 2\\
c)B = - {x^2} - 2x + 8\\
= - \left( {{x^2} + 2x - 8} \right)\\
= - \left( {{x^2} + 2x + 1 - 9} \right)\\
= - {\left( {x + 1} \right)^2} + 9\\
Do:{\left( {x + 1} \right)^2} \ge 0\forall x\\
\to - {\left( {x + 1} \right)^2} \le 0\\
\to - {\left( {x + 1} \right)^2} + 9 \le 9\\
\to Max = 9\\
\Leftrightarrow x = - 1
\end{array}\)
