Cho E = (3a + √9a – 3) / (a + √a -2) – (√a + 1) / (√a + 2) + (√a – 2) / (1 – √a)
1) Tìm ĐKXĐ của E
2) Rút gọn E
3) Tìm a để E < 0
4) Tính E khi a = 3+2√2
5) TÌm gt nguyên của a để E nguyên
6) Tìm gtnn của 1/E
Cảm ơn mn nhiều ạ (xin lỗi vì mình k biết viết phân số) ^_^
Đáp án:
$\begin{array}{l}
1)Dkxd:\left\{ \begin{array}{l}
a \ge 0\\
a \ne 1
\end{array} \right.\\
2)\\
E = \dfrac{{3a + \sqrt {9a} – 3}}{{a + \sqrt a – 2}} – \dfrac{{\sqrt a + 1}}{{\sqrt a + 2}} + \dfrac{{\sqrt a – 2}}{{1 – \sqrt a }}\\
= \dfrac{{3a + 3\sqrt a – 3}}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a – 1} \right)}} – \dfrac{{\sqrt a + 1}}{{\sqrt a + 2}} – \dfrac{{\sqrt a – 2}}{{\sqrt a – 1}}\\
= \dfrac{{3a + 3\sqrt a – 3 – \left( {\sqrt a + 1} \right)\left( {\sqrt a – 1} \right) – \left( {\sqrt a – 2} \right)\left( {\sqrt a + 2} \right)}}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a – 1} \right)}}\\
= \dfrac{{3a + 3\sqrt a – 3 – a + 1 – a + 4}}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a – 1} \right)}}\\
= \dfrac{{a + 3\sqrt a + 2}}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a – 1} \right)}}\\
= \dfrac{{\left( {\sqrt a + 2} \right)\left( {\sqrt a + 1} \right)}}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a – 1} \right)}}\\
= \dfrac{{\sqrt a + 1}}{{\sqrt a – 1}}\\
3)E < 0\\
\Rightarrow \dfrac{{\sqrt a + 1}}{{\sqrt a – 1}} < 0\\
\Rightarrow \sqrt a – 1 < 0\\
\Rightarrow \sqrt a < 1\\
\Rightarrow a < 1\\
Vay\,0 \le a < 1\\
4)a = 3 + 2\sqrt 2 \left( {tmdk} \right)\\
\Rightarrow a = {\left( {\sqrt 2 + 1} \right)^2}\\
\Rightarrow \sqrt a = \sqrt 2 + 1\\
\Rightarrow E = \dfrac{{\sqrt a + 1}}{{\sqrt a – 1}} = \dfrac{{\sqrt 2 + 1 + 1}}{{\sqrt 2 + 1 – 1}}\\
= \dfrac{{\sqrt 2 + 2}}{{\sqrt 2 }} = \sqrt 2 + 1\\
5)E = \dfrac{{\sqrt a + 1}}{{\sqrt a – 1}} = \dfrac{{\sqrt a – 1 + 2}}{{\sqrt a – 1}} = 1 + \dfrac{2}{{\sqrt a – 1}}\\
E \in Z\\
\Rightarrow \dfrac{2}{{\sqrt a – 1}} \in Z\\
\Rightarrow \sqrt a – 1 \in \left\{ { – 1;1;2} \right\}\left( {do:\sqrt a – 1 \ge – 1} \right)\\
\Rightarrow \sqrt a \in \left\{ {0;2;3} \right\}\\
\Rightarrow a \in \left\{ {0;4;9} \right\}\\
6)E = 1 + \dfrac{2}{{\sqrt a – 1}}\\
\sqrt a – 1 \ge – 1\\
\Rightarrow \dfrac{2}{{\sqrt a – 1}} \le \dfrac{2}{{ – 1}} = – 2\\
\Rightarrow 1 + \dfrac{2}{{\sqrt a – 1}} \le 1 – 2 = – 1\\
\Rightarrow E \le – 1\\
\Rightarrow \dfrac{1}{E} \ge – 1\\
\Rightarrow GTNN:\dfrac{1}{E} = – 1\,khi:x = 0
\end{array}$