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