Đáp án:
2) b) x>1
Giải thích các bước giải:
\(\begin{array}{l}
1)M = \dfrac{{\sqrt {12} + 4}}{{\sqrt 3 + \sqrt 5 }}:\sqrt 5 \\
= \dfrac{{2\sqrt 3 + 4}}{{\sqrt 3 + \sqrt 5 }}.\dfrac{1}{{\sqrt 5 }}\\
= \dfrac{{2\sqrt 3 + 4}}{{\sqrt {15} + 5}}\\
2)a)P = \left( {\dfrac{{1 - x}}{{x - \sqrt x + 1}}} \right):\dfrac{{x + 2\sqrt x + 1}}{{x\sqrt x + 1}}\\
= \dfrac{{ - \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{x - \sqrt x + 1}}.\dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{{{\left( {\sqrt x + 1} \right)}^2}}}\\
= - \left( {\sqrt x - 1} \right)\\
b)P < 0\\
\to - \left( {\sqrt x - 1} \right) < 0\\
\to \sqrt x - 1 > 0\\
\to x > 1
\end{array}\)
