Toán Lớp 9: Làm giúp em vs ạ, hứa cho ctlhn + 5*
P = ($\frac{\sqrt{x}}{\sqrt{x}-1}$ – $\frac{1}{x-\sqrt{x}}$) : ($\frac{1}{\sqrt{x}+1}$ + $\frac{2}{x-1}$)
a, Rút gọn P
b, Tính giá trị của x để P < 2
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Toán Lớp 9: Làm giúp em vs ạ, hứa cho ctlhn + 5*
P = ($\frac{\sqrt{x}}{\sqrt{x}-1}$ – $\frac{1}{x-\sqrt{x}}$) : ($\frac{1}{\sqrt{x}+1}$ + $\frac{2}{x-1}$)
a, Rút gọn P
b, Tính giá trị của x để P < 2
Comments ( 1 )
a,\\
P = \dfrac{{x – 1}}{{\sqrt x }}\\
b,\\
\left\{ \begin{array}{l}
0 < x < 3 + 2\sqrt 2 \\
x \ne 1
\end{array} \right.
\end{array}\)
DKXD:\,\,\,\left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.\\
a,\\
P = \left( {\dfrac{{\sqrt x }}{{\sqrt x – 1}} – \dfrac{1}{{x – \sqrt x }}} \right):\left( {\dfrac{1}{{\sqrt x + 1}} + \dfrac{2}{{x – 1}}} \right)\\
= \left( {\dfrac{{\sqrt x }}{{\sqrt x – 1}} – \dfrac{1}{{\sqrt x \left( {\sqrt x – 1} \right)}}} \right):\left( {\dfrac{1}{{\sqrt x + 1}} + \dfrac{2}{{{{\sqrt x }^2} – {1^2}}}} \right)\\
= \dfrac{{{{\sqrt x }^2} – 1}}{{\sqrt x \left( {\sqrt x – 1} \right)}}:\left( {\dfrac{1}{{\sqrt x + 1}} + \dfrac{2}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}} \right)\\
= \dfrac{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x – 1} \right)}}:\dfrac{{\left( {\sqrt x – 1} \right) + 2}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x }}:\dfrac{{\sqrt x + 1}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x }}:\dfrac{1}{{\sqrt x – 1}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x – 1} \right)}}{{\sqrt x }}\\
= \dfrac{{{{\sqrt x }^2} – {1^2}}}{{\sqrt x }}\\
= \dfrac{{x – 1}}{{\sqrt x }}\\
b,\\
P < 2 \Leftrightarrow \dfrac{{x – 1}}{{\sqrt x }} < 2\\
\Leftrightarrow x – 1 < 2\sqrt x \\
\Leftrightarrow x – 2\sqrt x – 1 < 0\\
\Leftrightarrow x – 2\sqrt x + 1 < 2\\
\Leftrightarrow {\sqrt x ^2} – 2.\sqrt x .1 + {1^2} < 2\\
\Leftrightarrow {\left( {\sqrt x – 1} \right)^2} < 2\\
\Leftrightarrow – \sqrt 2 < \sqrt x – 1 < \sqrt 2 \\
\Leftrightarrow – \sqrt 2 + 1 < \sqrt x < \sqrt 2 + 1\\
x > 0 \Rightarrow \sqrt x > 0\\
\Rightarrow 0 < \sqrt x < \sqrt 2 + 1\\
\Leftrightarrow 0 < x < {\left( {\sqrt 2 + 1} \right)^2}\\
\Leftrightarrow 0 < x < 3 + 2\sqrt 2 \\
\Rightarrow \left\{ \begin{array}{l}
0 < x < 3 + 2\sqrt 2 \\
x \ne 1
\end{array} \right.
\end{array}\)