Đáp án:
b) B=2
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:x = \dfrac{1}{9}\\
\to A = \dfrac{{\dfrac{1}{9} + 3}}{{\sqrt {\dfrac{1}{9}} + 1}} = \dfrac{7}{3}\\
b)B = \dfrac{{2{{\left( {x + 1} \right)}^2}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} + \dfrac{2}{{1 - \sqrt x }} + \dfrac{2}{{\sqrt x + 1}}\\
= \dfrac{{2\left( {x + 1} \right)}}{{x - 1}} - \dfrac{2}{{\sqrt x - 1}} + \dfrac{2}{{\sqrt x + 1}}\\
= \dfrac{{2x + 2 - 2\left( {\sqrt x + 1} \right) + 2\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{2x + 2 - 2\sqrt x - 2 + 2\sqrt x - 2}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{2x - 2}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}} = \dfrac{{2\left( {x - 1} \right)}}{{x - 1}}\\
= 2\\
3)A \le B\\
\to \dfrac{{x + 3}}{{\sqrt x + 1}} \le 2\\
\to \dfrac{{x + 3 - 2\sqrt x - 2}}{{\sqrt x + 1}} \le 0\\
\to \dfrac{{x - 2\sqrt x + 1}}{{\sqrt x + 1}} \le 0\\
\to x - 2\sqrt x + 1 \le 0\left( {do:\sqrt x + 1 > 0\forall x \ge 0} \right)\\
\to {\left( {\sqrt x - 1} \right)^2} \le 0\\
\Leftrightarrow \sqrt x - 1 = 0\\
\to x = 1\left( {KTM} \right)\\
\to x \in \emptyset
\end{array}\)
