Toán Lớp 9: cho biết A=√x/√x+3 + 2√x/√x-3 – 3x+9/x-9 (x≥0; x khác 9)
a. rút gọn A
b.Tìm x ,để A< 1/6
c. tìm x thuộc Z, để A thuộc Z
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Toán Lớp 9: cho biết A=√x/√x+3 + 2√x/√x-3 – 3x+9/x-9 (x≥0; x khác 9)
a. rút gọn A
b.Tìm x ,để A< 1/6
c. tìm x thuộc Z, để A thuộc Z
Comments ( 1 )
Dkxd:x \ge 0;x \ne 9\\
a)A = \dfrac{{\sqrt x }}{{\sqrt x + 3}} + \dfrac{{2\sqrt x }}{{\sqrt x – 3}} – \dfrac{{3x + 9}}{{x – 9}}\\
= \dfrac{{\sqrt x \left( {\sqrt x – 3} \right) + 2\sqrt x \left( {\sqrt x + 3} \right) – 3x – 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}\\
= \dfrac{{x – 3\sqrt x + 2x + 6\sqrt x – 3x – 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}\\
= \dfrac{{3\sqrt x – 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}\\
= \dfrac{{3\left( {\sqrt x – 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}\\
= \dfrac{3}{{\sqrt x + 3}}\\
b)A < \dfrac{1}{6}\\
\Leftrightarrow \dfrac{3}{{\sqrt x + 3}} – \dfrac{1}{6} < 0\\
\Leftrightarrow \dfrac{{18 – \sqrt x – 3}}{{6\left( {\sqrt x + 3} \right)}} < 0\\
\Leftrightarrow 15 – \sqrt x < 0\\
\Leftrightarrow \sqrt x > 15\\
\Leftrightarrow x > 225\\
Vậy\,x > 225\\
c)A = \dfrac{3}{{\sqrt x + 3}} \in Z\\
\Leftrightarrow \sqrt x + 3 = 3\left( {do:\sqrt x + 3 \ge 3} \right)\\
\Leftrightarrow \sqrt x = 0\\
\Leftrightarrow x = 0\\
Vậy\,x = 0
\end{array}$