($\frac{2x + 1}{x√x – 1}$ – $\frac{1}{√x – 1}$ ) : ( 1 – $\frac{x – 2}{x + √x + 1 }$ )
a rút gọn b) tính A biết x = 2 – √3 /2 c) tìm x thuộc z đẻ a thuộc z d) tìm gtnn của A e) so sanh a với 1 f) tìm x để A > 1/2 g ) tìm x để A =1/3
($\frac{2x + 1}{x√x – 1}$ – $\frac{1}{√x – 1}$ ) : ( 1 – $\frac{x – 2}{x + √x + 1 }$ ) a rút gọn b) tính A biết x = 2 – √3 /2 c) tìm x thuộc z đ
By Adalynn
Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x \ne 1\\
a)A = \left( {\frac{{2x + 1}}{{x\sqrt x – 1}} – \frac{1}{{\sqrt x – 1}}} \right):\left( {1 – \frac{{x – 2}}{{x + \sqrt x + 1}}} \right)\\
= \frac{{2x + 1 – \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}}:\left( {\frac{{x + \sqrt x + 1 – x + 2}}{{x + \sqrt x + 1}}} \right)\\
= \frac{{2x + 1 – x – \sqrt x – 1}}{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}}.\frac{{x + \sqrt x + 1}}{{\sqrt x + 3}}\\
= \frac{{x – \sqrt x }}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \frac{{\sqrt x }}{{\sqrt x + 3}}\\
b)x = \frac{{2 – \sqrt 3 }}{2}\left( {tmdk} \right)\\
x = \frac{{4 – 2\sqrt 3 }}{4} = \frac{{{{\left( {\sqrt 3 – 1} \right)}^2}}}{{{2^2}}} = {\left( {\frac{{\sqrt 3 – 1}}{2}} \right)^2}\\
\Rightarrow \sqrt x = \frac{{\sqrt 3 – 1}}{2} = \frac{{3 – 1}}{{2\left( {\sqrt 3 + 1} \right)}} = \frac{1}{{\sqrt 3 + 1}}\\
\Rightarrow A = \frac{{\sqrt x }}{{\sqrt x + 3}} = \frac{{\frac{1}{{\sqrt 3 + 1}}}}{{\frac{1}{{\sqrt 3 + 1}} + 3}} = \frac{1}{{1 + 3\sqrt 3 + 3}} = \frac{{3\sqrt 3 – 4}}{{{{\left( {3\sqrt 3 } \right)}^2} – {4^2}}} = \frac{{3\sqrt 3 – 4}}{9}\\
c)A = \frac{{\sqrt x }}{{\sqrt x + 3}} = 1 – \frac{3}{{\sqrt x + 3}}\\
A \in Z\\
\Rightarrow \frac{3}{{\sqrt x + 3}} \in Z\\
\Rightarrow \left( {\sqrt x + 3} \right) \in U\left( 3 \right)\\
Do:\sqrt x + 3 \ge 3\forall x \ge 0\\
\Rightarrow \sqrt x + 3 = 3\\
\Rightarrow x = 0\left( {tmdk} \right)\\
d)A = 1 – \frac{3}{{\sqrt x + 3}}\\
Do:\sqrt x + 3 \ge 3\\
\Rightarrow \frac{3}{{\sqrt x + 3}} \le \frac{3}{3} = 1\\
\Rightarrow – \frac{3}{{\sqrt x + 3}} \ge – 1\\
\Rightarrow 1 – \frac{3}{{\sqrt x + 3}} \ge 0\\
\Rightarrow GTNN:A = 0 \Leftrightarrow x = 0\\
e)A = \frac{{\sqrt x }}{{\sqrt x + 3}} < 1\left( {do:\sqrt x \ge 0} \right)\\
f)A > \frac{1}{2}\\
\Rightarrow \frac{{\sqrt x }}{{\sqrt x + 3}} > \frac{1}{2}\\
\Rightarrow 2\sqrt x > \sqrt x + 3\\
\Rightarrow \sqrt x > 3\\
\Rightarrow x > 9\\
g)A = \frac{1}{3}\\
\Rightarrow \frac{{\sqrt x }}{{\sqrt x + 3}} = \frac{1}{3}\\
\Rightarrow 3\sqrt x = \sqrt x + 3\\
\Rightarrow 2\sqrt x = 3\\
\Rightarrow \sqrt x = \frac{3}{2}\\
\Rightarrow x = \frac{9}{4}\left( {tmdk} \right)
\end{array}$