Đáp án:
\(\begin{array}{l}
a)\,P = \dfrac{{x - 1}}{{\sqrt x }}\\
b)\,0 < x < 1\\
c)\,P = \dfrac{{ - 3 + \sqrt 3 }}{2}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\,DK:\,x > 0;x \ne 1\\
P = \dfrac{{\sqrt x .\sqrt x - 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}:\dfrac{{\sqrt x - 1 + 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}\\
= \dfrac{{x - 1}}{{\sqrt x }}\\
b)\,P < 0 \Leftrightarrow \dfrac{{x - 1}}{{\sqrt x }} < 0\\
\Rightarrow x - 1 < 0\,\left( {do\,\,\sqrt x > 0} \right)\\
\Leftrightarrow x < 1\\
KHDK \Rightarrow 0 < x < 1\\
c)\,x = 4 - 2\sqrt 3 = {\left( {\sqrt 3 - 1} \right)^2}\\
\Rightarrow \sqrt x = \sqrt 3 - 1\\
\Rightarrow P = \dfrac{{4 - 2\sqrt 3 - 1}}{{\sqrt 3 - 1}} = \dfrac{{3 - 2\sqrt 3 }}{{\sqrt 3 - 1}}\\
= \dfrac{{ - 3 + \sqrt 3 }}{2}
\end{array}\)
