Đáp án:
a) MaxB=8
Giải thích các bước giải:
\(\begin{array}{l}
a)B = - \left( {x - 6\sqrt x + 1} \right)\\
= - \left( {x - 6\sqrt x + 9 - 8} \right)\\
= - {\left( {\sqrt x - 3} \right)^2} + 8\\
Do:{\left( {\sqrt x - 3} \right)^2} \ge 0\forall x \ge 0\\
\to - {\left( {\sqrt x - 3} \right)^2} \le 0\\
\to - {\left( {\sqrt x - 3} \right)^2} + 8 \le 8\\
\to Max = 8\\
\Leftrightarrow x = 9\\
C = \dfrac{1}{{x - \sqrt x - 1}} = \dfrac{1}{{x - 2\sqrt x .\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{5}{4}}}\\
= \dfrac{1}{{{{\left( {\sqrt x - \dfrac{1}{2}} \right)}^2} - \dfrac{5}{4}}}\\
Do:{\left( {\sqrt x - \dfrac{1}{2}} \right)^2} \ge 0\forall x \ge 0\\
\to {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} - \dfrac{5}{4} \ge - \dfrac{5}{4}\\
\to \dfrac{1}{{{{\left( {\sqrt x - \dfrac{1}{2}} \right)}^2} - \dfrac{5}{4}}} \le - \dfrac{4}{5}\\
\to Max = - \dfrac{4}{5}\\
\Leftrightarrow x = \dfrac{1}{4}\\
b)A = \dfrac{{x + 2}}{{x - 5}} = \dfrac{{x - 5 + 7}}{{x - 5}} = 1 + \dfrac{7}{{x - 5}}\\
A \in Z \Leftrightarrow \dfrac{7}{{x - 5}} \in Z\\
\Leftrightarrow x - 5 \in U\left( 7 \right)\\
\to \left[ \begin{array}{l}
x - 5 = 7\\
x - 5 = - 7\\
x - 5 = 1\\
x - 5 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 12\\
x = - 2\\
x = 6\\
x = 4
\end{array} \right.\\
B = \dfrac{{3x + 1}}{{2 - x}} = \dfrac{{ - 3\left( { - x + 2} \right) + 7}}{{2 - x}}\\
= - 3 + \dfrac{7}{{2 - x}}\\
B \in Z \Leftrightarrow \dfrac{7}{{2 - x}} \in Z\\
\Leftrightarrow 2 - x \in U\left( 7 \right)\\
\to \left[ \begin{array}{l}
2 - x = 7\\
2 - x = - 7\\
2 - x = 1\\
2 - x = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = - 5\\
x = 9\\
x = 1\\
x = 3
\end{array} \right.
\end{array}\)
( câu A phần a thiếu đề bạn )
