Đáp án:
\(\begin{array}{l}
a)A = 1\\
B = \dfrac{{3\sqrt x }}{{\sqrt x + 2}}\\
b)0 \le x < 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \left( {3.4\sqrt 2 - 2.3\sqrt 2 - 5\sqrt 2 } \right):\sqrt 2 \\
= \dfrac{{\sqrt 2 }}{{\sqrt 2 }} = 1\\
B = \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right) + 2\sqrt x \left( {\sqrt x - 2} \right) - 5\sqrt x - 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x + 3\sqrt x + 2 + 2x - 4\sqrt x - 5\sqrt x - 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{3x - 6\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{3\sqrt x }}{{\sqrt x + 2}}\\
b)A > B\\
\to 1 > \dfrac{{3\sqrt x }}{{\sqrt x + 2}}\\
\to \dfrac{{3\sqrt x - \sqrt x - 2}}{{\sqrt x + 2}} < 0\\
\to 2\sqrt x - 2 < 0\left( {do:\sqrt x + 2 > 0\forall x \ge 0} \right)\\
\to 0 \le x < 1
\end{array}\)
