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
1 câu trả lời
Đáp án:
\(\begin{array}{l}
a,\\
P = \dfrac{{x - 1}}{{\sqrt x }}\\
b,\\
\left\{ \begin{array}{l}
0 < x < 3 + 2\sqrt 2 \\
x \ne 1
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
DKXD:\,\,\,\left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.\\
a,\\
P = \left( {\dfrac{{\sqrt x }}{{\sqrt x - 1}} - \dfrac{1}{{x - \sqrt x }}} \right):\left( {\dfrac{1}{{\sqrt x + 1}} + \dfrac{2}{{x - 1}}} \right)\\
= \left( {\dfrac{{\sqrt x }}{{\sqrt x - 1}} - \dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}}} \right):\left( {\dfrac{1}{{\sqrt x + 1}} + \dfrac{2}{{{{\sqrt x }^2} - {1^2}}}} \right)\\
= \dfrac{{{{\sqrt x }^2} - 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}:\left( {\dfrac{1}{{\sqrt x + 1}} + \dfrac{2}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right)\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}}:\dfrac{{\left( {\sqrt x - 1} \right) + 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x }}:\dfrac{{\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x }}:\dfrac{1}{{\sqrt x - 1}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{\sqrt x }}\\
= \dfrac{{{{\sqrt x }^2} - {1^2}}}{{\sqrt x }}\\
= \dfrac{{x - 1}}{{\sqrt x }}\\
b,\\
P < 2 \Leftrightarrow \dfrac{{x - 1}}{{\sqrt x }} < 2\\
\Leftrightarrow x - 1 < 2\sqrt x \\
\Leftrightarrow x - 2\sqrt x - 1 < 0\\
\Leftrightarrow x - 2\sqrt x + 1 < 2\\
\Leftrightarrow {\sqrt x ^2} - 2.\sqrt x .1 + {1^2} < 2\\
\Leftrightarrow {\left( {\sqrt x - 1} \right)^2} < 2\\
\Leftrightarrow - \sqrt 2 < \sqrt x - 1 < \sqrt 2 \\
\Leftrightarrow - \sqrt 2 + 1 < \sqrt x < \sqrt 2 + 1\\
x > 0 \Rightarrow \sqrt x > 0\\
\Rightarrow 0 < \sqrt x < \sqrt 2 + 1\\
\Leftrightarrow 0 < x < {\left( {\sqrt 2 + 1} \right)^2}\\
\Leftrightarrow 0 < x < 3 + 2\sqrt 2 \\
\Rightarrow \left\{ \begin{array}{l}
0 < x < 3 + 2\sqrt 2 \\
x \ne 1
\end{array} \right.
\end{array}\)