Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
B = 5x - 3{x^2} + 6\\
= - 3\left( {{x^2} - \frac{5}{3}x + \frac{{25}}{{36}}} \right) + \frac{{97}}{{12}}\\
= - 3{\left( {x - \frac{5}{6}} \right)^2} + \frac{{97}}{{12}} \le \frac{{97}}{{12}}\\
= > B \le \frac{{97}}{{12}}
\end{array}\]

dấu = xảy ra<=>\[x - \frac{5}{6} = 0 \Leftrightarrow x = \frac{5}{6}\]

vậy maxB =\[\frac{{97}}{{12}}\] đạt được khi \[x = \frac{5}{6}\]

\[\begin{array}{l}
C = - 4{x^2} + x - 1\\
= - 4\left( {{x^2} - \frac{1}{4}x + \frac{1}{{64}}} \right) - \frac{{15}}{{16}}\\
= - 4{\left( {x - \frac{1}{8}} \right)^2} - \frac{{15}}{{16}} \le \frac{{ - 15}}{{16}}\\
= > C \le \frac{{ - 15}}{{16}}
\end{array}\]

dấu = xảy ra<=>\[x - \frac{1}{8} = 0 \Leftrightarrow x = \frac{1}{8}\]

Vậy maxC=\[\frac{{ - 15}}{{16}}\] đạt được khi \[x = \frac{1}{8}\]

\[\begin{array}{l}
D = {x^2} - x + 1\\
= {x^2} - x + \frac{1}{4} + \frac{3}{4}\\
= {\left( {x - \frac{1}{2}} \right)^2} + \frac{3}{4} \ge \frac{3}{4}\\
= > D \ge \frac{3}{4}
\end{array}\]

dấu = xảy ra<=>\[x - \frac{1}{2} = 0 \Leftrightarrow x = \frac{1}{2}\]

vậy maxD=\[\frac{3}{4}\] đạt được khi \[x = \frac{1}{2}\]

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Answer 2