Đáp án:
c) \(0 < x < 25;x \ne 1\)
Giải thích các bước giải:
\(\begin{array}{l}
b)B = \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right) + \sqrt x - 1 + 5 - 2\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x - 4 - \sqrt x + 4}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 2}}\\
c)A:B < 4\\
\to \dfrac{{4\sqrt x }}{{\sqrt x - 5}}:\dfrac{{\sqrt x }}{{\sqrt x + 2}} < 4\\
\to \dfrac{{4\sqrt x }}{{\sqrt x - 5}}.\dfrac{{\sqrt x + 2}}{{\sqrt x }} < 4\\
\to \dfrac{{4\left( {\sqrt x + 2} \right)}}{{\sqrt x - 5}} < 4\\
\to \dfrac{{\sqrt x + 2}}{{\sqrt x - 5}} < 1\\
\to \dfrac{{\sqrt x + 2 - \sqrt x + 5}}{{\sqrt x - 5}} < 0\\
\to \dfrac{7}{{\sqrt x - 5}} < 0\\
\to \sqrt x - 5 < 0\\
\to 0 < x < 25;x \ne 1
\end{array}\)
