Toán Lớp 9: P = [(x + 2)/(x\sqrt{x} -1) + (\sqrt{x})/(x + \sqrt{x} + 1) – 1/(\sqrt{x} – 1)] : (\sqrt{x – 1})/2
Đk : x >= 0 ; x khác 4
a) Rút gọn
b) Tìm x để P = 2/7
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Toán Lớp 9: P = [(x + 2)/(x\sqrt{x} -1) + (\sqrt{x})/(x + \sqrt{x} + 1) – 1/(\sqrt{x} – 1)] : (\sqrt{x – 1})/2
Đk : x >= 0 ; x khác 4
a) Rút gọn
b) Tìm x để P = 2/7
Comments ( 1 )
Dkxd:x \ge 0;x \ne 1\\
a)P = \left( {\dfrac{{x + 2}}{{x\sqrt x – 1}} + \dfrac{{\sqrt x }}{{x + \sqrt x + 1}} – \dfrac{1}{{\sqrt x – 1}}} \right):\dfrac{{\sqrt x – 1}}{2}\\
= \dfrac{{x + 2 + \sqrt x \left( {\sqrt x – 1} \right) – \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{2}{{\sqrt x – 1}}\\
= \dfrac{{x + 2 + x – \sqrt x – x – \sqrt x – 1}}{{{{\left( {\sqrt x – 1} \right)}^2}\left( {x + \sqrt x + 1} \right)}}.2\\
= 2.\dfrac{{x – 2\sqrt x + 1}}{{{{\left( {\sqrt x – 1} \right)}^2}\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{2.{{\left( {\sqrt x – 1} \right)}^2}}}{{{{\left( {\sqrt x – 1} \right)}^2}\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{2}{{x + \sqrt x + 1}}\\
b)P = \dfrac{2}{7}\\
\Leftrightarrow \dfrac{2}{{x + \sqrt x + 1}} = \dfrac{2}{7}\\
\Leftrightarrow x + \sqrt x + 1 = 7\\
\Leftrightarrow x + \sqrt x – 6 = 0\\
\Leftrightarrow \left( {\sqrt x – 2} \right)\left( {\sqrt x + 3} \right) = 0\\
\Leftrightarrow \sqrt x = 2\\
\Leftrightarrow x = 4\left( {tmdk} \right)\\
Vậy\,x = 4
\end{array}$