Đáp án:
c) \(1 > x \ge 0\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:x = 16\\
\to B = \sqrt {16} + 1 = 5\\
b)A = \left[ {\dfrac{{x + \sqrt x + 10 - \sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}} \right].\left( {\sqrt x - 3} \right)\\
= \dfrac{{x + 7}}{{\sqrt x + 3}}\\
c)A > B\\
\to \dfrac{{x + 7}}{{\sqrt x + 3}} > \sqrt x + 1\\
\to \dfrac{{x + 7 - \left( {\sqrt x + 1} \right)\left( {\sqrt x + 3} \right)}}{{\sqrt x + 3}} > 0\\
\to x + 7 - \left( {\sqrt x + 1} \right)\left( {\sqrt x + 3} \right) > 0\left( {do:\sqrt x + 3 > 0\forall x \ge 0} \right)\\
\to x + 7 - x - 4\sqrt x - 3 > 0\\
\to 4 - 4\sqrt x > 0\\
\to 1 > x\\
\to 1 > x \ge 0
\end{array}\)
