Cho biểu thức M= $\frac{2√x-9}{x-5√x+6}$ – $\frac{√x+3}{√x-2}$ – $\frac{2√x}{3-√x}$ (x $\geq$ 0: x $\neq$ 9; x $\neq$ 4)
1) rút gọn M
2) tìm x để M ² = 3M
3) Tìm x để M là số nguyên
4) tìm x để M < 1
5) tìm x để A = $\frac{1}{M}$ giá trị nhỏ nhất
Đáp án:
1)\(M = \dfrac{{\sqrt x }}{{\sqrt x – 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)M = \dfrac{{2\sqrt x – 9 – \left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right) + 2\sqrt x \left( {\sqrt x – 2} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 2} \right)}}\\
= \dfrac{{2\sqrt x – 9 – x + 9 + 2x – 4\sqrt x }}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 2} \right)}}\\
= \dfrac{{x – 2\sqrt x }}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 2} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x – 3}}\\
2){M^2} = 3M\\
\to {M^2} – 3M = 0\\
\to M\left( {M – 3} \right) = 0\\
\to \left[ \begin{array}{l}
M = 0\\
M = 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
\dfrac{{\sqrt x }}{{\sqrt x – 3}} = 0\\
\dfrac{{\sqrt x }}{{\sqrt x – 3}} = 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
\sqrt x = 3\sqrt x – 9
\end{array} \right.\\
\to \left[ \begin{array}{l}
2\sqrt x = 9\\
x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{81}}{4}\\
x = 0
\end{array} \right.\left( {TM} \right)\\
3)M = \dfrac{{\sqrt x }}{{\sqrt x – 3}} = \dfrac{{\sqrt x – 3 + 3}}{{\sqrt x – 3}} = 1 + \dfrac{3}{{\sqrt x – 3}}\\
M \in Z \Leftrightarrow \dfrac{3}{{\sqrt x – 3}} \in Z\\
\to \sqrt x – 3 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
\sqrt x – 3 = 3\\
\sqrt x – 3 = – 3\\
\sqrt x – 3 = 1\\
\sqrt x – 3 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 36\\
x = 0\\
x = 16\\
x = 4\left( l \right)
\end{array} \right.\\
4)M < 1\\
\to 1 + \dfrac{3}{{\sqrt x – 3}} < 1\\
\to \dfrac{3}{{\sqrt x – 3}} < 0\\
\to \sqrt x – 3 < 0\\
\to x < 9\\
\to 0 \le x < 9;x \ne 4\\
5)A = \dfrac{1}{M} = \dfrac{{\sqrt x – 3}}{{\sqrt x }} = 1 – \dfrac{3}{{\sqrt x }}\\
DK:x \ge 1\\
\to \sqrt x \ge 1\\
\to \dfrac{3}{{\sqrt x }} \le 3\\
\to – \dfrac{3}{{\sqrt x }} \ge – 3\\
\to 1 – \dfrac{3}{{\sqrt x }} \ge – 2\\
\to Min = – 2\\
\Leftrightarrow x = 1
\end{array}\)