Toán Lớp 9: Cho biểu thức A=2√x -5/√x+2 ( x lớn hơn hoặc = 0, x khác 4)
a) tìm giá trị của A tại x = 1/9
b) Tìm x để A=1,A=2/3,A= -1/2
c) Tìm x để A bé hơn hoặc = -2/3, A > 1/5
d) So sánh A với 2
e) Tìm giá trị nguyên của x để A có giá trị nguyên.
Giúp với ạ????
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Comments ( 2 )
x \ge 0;x\# 4\\
a)x = \dfrac{1}{9}\left( {tmdk} \right)\\
\Leftrightarrow \sqrt x = \dfrac{1}{3}\\
\Leftrightarrow A = \dfrac{{2\sqrt x – 5}}{{\sqrt x + 2}} = \dfrac{{2.\dfrac{1}{3} – 5}}{{\dfrac{1}{3} + 2}} = \dfrac{{ – 13}}{7}\\
b)A = 1\\
\Leftrightarrow \dfrac{{2\sqrt x – 5}}{{\sqrt x + 2}} = 1\\
\Leftrightarrow 2\sqrt x – 5 = \sqrt x + 2\\
\Leftrightarrow \sqrt x = 7\\
\Leftrightarrow x = 49\left( {tmdk} \right)\\
A = \dfrac{2}{3}\\
\Leftrightarrow \dfrac{{2\sqrt x – 5}}{{\sqrt x + 2}} = \dfrac{2}{3}\\
\Leftrightarrow 6\sqrt x – 15 = 2\sqrt x + 4\\
\Leftrightarrow 4\sqrt x = 19\\
\Leftrightarrow \sqrt x = \dfrac{{19}}{4}\\
\Leftrightarrow x = \dfrac{{361}}{{16}}\left( {tmdk} \right)\\
A = \dfrac{{ – 1}}{2}\\
\Leftrightarrow \dfrac{{2\sqrt x – 5}}{{\sqrt x + 2}} = – \dfrac{1}{2}\\
\Leftrightarrow 4\sqrt x – 10 = – \sqrt x – 2\\
\Leftrightarrow 5\sqrt x = 8\\
\Leftrightarrow \sqrt x = \dfrac{8}{5}\\
\Leftrightarrow x = \dfrac{{64}}{{25}}\left( {tmdk} \right)\\
c)A \le – \dfrac{2}{3}\\
\Leftrightarrow \dfrac{{2\sqrt x – 5}}{{\sqrt x + 2}} \le \dfrac{{ – 2}}{3}\\
\Leftrightarrow \dfrac{{2\sqrt x – 5}}{{\sqrt x + 2}} + \dfrac{2}{3} \le 0\\
\Leftrightarrow \dfrac{{6\sqrt x – 15 + 2\sqrt x + 4}}{{3\left( {\sqrt x + 2} \right)}} \le 0\\
\Leftrightarrow 8\sqrt x – 11 \le 0\\
\Leftrightarrow \sqrt x \le \dfrac{{11}}{8}\\
\Leftrightarrow x \le \dfrac{{121}}{{64}}\\
Vậy\,0 \le x \le \dfrac{{121}}{{64}}\\
A > \dfrac{1}{5}\\
\Leftrightarrow \dfrac{{2\sqrt x – 5}}{{\sqrt x + 2}} > \dfrac{1}{5}\\
\Leftrightarrow \dfrac{{10\sqrt x – 25 – \sqrt x – 2}}{{5\left( {\sqrt x + 2} \right)}} > 0\\
\Leftrightarrow 9\sqrt x – 27 > 0\\
\Leftrightarrow \sqrt x > 3\\
\Leftrightarrow x > 9\\
Vậy\,x > 9\\
d)A – 2\\
= \dfrac{{2\sqrt x – 5}}{{\sqrt x + 2}} – 2\\
= \dfrac{{2\sqrt x – 5 – 2\sqrt x – 4}}{{\sqrt x + 2}}\\
= \dfrac{{ – 9}}{{\sqrt x + 2}} < 0\\
\Leftrightarrow A < 2\\
e)A = \dfrac{{2\sqrt x – 5}}{{\sqrt x + 2}}\\
= \dfrac{{2\sqrt x + 4 – 9}}{{\sqrt x + 2}}\\
= 2 – \dfrac{9}{{\sqrt x + 2}}\\
\Leftrightarrow \dfrac{9}{{\sqrt x + 2}} \in Z\\
\Leftrightarrow \sqrt x + 2 \in \left\{ {3;9} \right\}\\
\Leftrightarrow \sqrt x \in \left\{ {1;7} \right\}\\
\Leftrightarrow x \in \left\{ {1;49} \right\}\left( {tmdk} \right)\\
Vậy\,x \in \left\{ {1;49} \right\}
\end{array}$