Chậm thì còn cái nịt nhé~
`A=(\frac{2\sqrtx }{\sqrtx+3}+\frac{\sqrtx}{\sqrtx-3}-\frac{3x+3}{x-9}):(\frac{2\sqrtx-2}{\sqrtx-3}-1)`
a) Tìm ĐK và rút gọn A
b) Tìm x để `A<\frac{-1}{2}`
c) Tìm min A
Chậm thì còn cái nịt nhé~
`A=(\frac{2\sqrtx }{\sqrtx+3}+\frac{\sqrtx}{\sqrtx-3}-\frac{3x+3}{x-9}):(\frac{2\sqrtx-2}{\sqrtx-3}-1)`
a) Tìm ĐK và rút gọn A
b) Tìm x để `A<\frac{-1}{2}`
c) Tìm min A
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a) Điều kiện xác định:
$\begin{array}{l} \left\{ \begin{array}{l} x \ge 0\\ \sqrt x – 3 \ne 0\\ \dfrac{{2\sqrt x – 2}}{{\sqrt x – 3}} – 1 \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ge 0\\ x \ne 9\\ \dfrac{{2\sqrt x – 2 – \sqrt x + 3}}{{\sqrt x – 3}} \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ge 0\\ x \ne 9\\ \dfrac{{\sqrt x + 1}}{{\sqrt x – 3}} \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x \ge 0\\ x \ne 9 \end{array} \right.\left( {do\,\sqrt x + 1 > 0} \right) \end{array}$
b)
$\begin{array}{l} A = \left( {\dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{\sqrt x – 3}} – \dfrac{{3x + 3}}{{x – 9}}} \right):\left( {\dfrac{{2\sqrt x – 2}}{{\sqrt x – 3}} – 1} \right)\\ A = \dfrac{{2\sqrt x \left( {\sqrt x – 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) – 3x – 3}}{{x – 9}}:\dfrac{{\sqrt x + 1}}{{\sqrt x – 3}}\\ A = \dfrac{{2x – 6\sqrt x + x + 3\sqrt x – 3x – 3}}{{x – 9}}.\dfrac{{\sqrt x – 3}}{{\sqrt x + 1}}\\ A = \dfrac{{ – 3\sqrt x – 3}}{{x – 9}}.\dfrac{{\sqrt x – 3}}{{\sqrt x + 1}} = \dfrac{{ – 3\left( {\sqrt x + 1} \right)\left( {\sqrt x – 3} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)\left( {\sqrt x + 1} \right)}}\\ A = \dfrac{{ – 3}}{{\sqrt x + 3}} < – \dfrac{1}{2}\\ \Rightarrow \dfrac{3}{{\sqrt x + 3}} > \dfrac{1}{2} \Leftrightarrow 6 > \sqrt x + 3\\ \Leftrightarrow \sqrt x < 3 \Leftrightarrow x < 9 \Rightarrow 0 \le x < 9\\ c)\\ A = \dfrac{{ – 3}}{{\sqrt x + 3}}\min \Rightarrow \sqrt x + 3\min \Rightarrow \min \sqrt x + 3 = 3 \Rightarrow x = 0\\ \Rightarrow \min A = – 1 \end{array}$