Đáp án:
\(\begin{array}{l}
1)A = \dfrac{5}{3}\\
2)\dfrac{{2\sqrt x - 9}}{{\sqrt x + 2}}
\end{array}\)
3) \(\left[ \begin{array}{l}
x = 9\\
x = \dfrac{1}{4}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)Thay:x = 16\\
\to A = \dfrac{{3\sqrt {16} - 2}}{{\sqrt {16} + 2}} = \dfrac{{10}}{6} = \dfrac{5}{3}\\
2)B = \dfrac{{2x - 2\left( {\sqrt x + 2} \right) - 7\sqrt x + 4}}{{\sqrt x \left( {\sqrt x + 2} \right)}}\\
= \dfrac{{2x - 2\sqrt x - 4 - 7\sqrt x + 4}}{{\sqrt x \left( {\sqrt x + 2} \right)}}\\
= \dfrac{{2x - 9\sqrt x }}{{\sqrt x \left( {\sqrt x + 2} \right)}}\\
= \dfrac{{2\sqrt x - 9}}{{\sqrt x + 2}}\\
3)M = A - B = \dfrac{{3\sqrt x - 2}}{{\sqrt x + 2}} - \dfrac{{2\sqrt x - 9}}{{\sqrt x + 2}}\\
= \dfrac{{\sqrt x + 7}}{{\sqrt x + 2}} = \dfrac{{\sqrt x + 2 + 5}}{{\sqrt x + 2}}\\
= 1 + \dfrac{5}{{\sqrt x + 2}}\\
\end{array}\)
\(\begin{array}{l}
Do:\left\{ \begin{array}{l}
\sqrt x + 7 > 0\\
\sqrt x + 2 > 0
\end{array} \right.\forall x > 0\\
\to \dfrac{{\sqrt x + 7}}{{\sqrt x + 2}} > 0\\
Do:\sqrt x > 0\forall x > 0\\
\to \sqrt x + 2 > 2\\
\to \dfrac{5}{{\sqrt x + 2}} < \dfrac{5}{2}\\
\to 1 + \dfrac{5}{{\sqrt x + 2}} < \dfrac{7}{2}\\
\to \dfrac{{\sqrt x + 7}}{{\sqrt x + 2}} < \dfrac{7}{2}\\
\to 0 < M < \dfrac{7}{2}\\
\to M \in \left\{ {1;2;3} \right\}\\
\to \left[ \begin{array}{l}
M = 1\\
M = 2\\
M = 3
\end{array} \right. \to \left[ \begin{array}{l}
\dfrac{{\sqrt x + 7}}{{\sqrt x + 2}} = 1\\
\dfrac{{\sqrt x + 7}}{{\sqrt x + 2}} = 2\\
\dfrac{{\sqrt x + 7}}{{\sqrt x + 2}} = 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x + 7 = \sqrt x + 2\left( l \right)\\
\sqrt x + 7 = 2\left( {\sqrt x + 2} \right)\\
\sqrt x + 7 = 3\left( {\sqrt x + 2} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = 3\\
2\sqrt x = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 9\\
x = \dfrac{1}{4}
\end{array} \right.
\end{array}\)
