Đáp án:
c) \(0 < x < 9;x \ne 1\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:x = 25\\
Q = \dfrac{{\sqrt {25} - 1}}{{\sqrt {25} + 1}} = \dfrac{4}{6} = \dfrac{2}{3}\\
b)DK:x > 0;x \ne 1\\
P = \dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}} + \dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x + 1} \right)}} - \dfrac{4}{{\sqrt x }}\\
= \dfrac{{x + \sqrt x + 1 + x - \sqrt x + 1 - 4}}{{\sqrt x }}\\
= \dfrac{{2x - 2}}{{\sqrt x }}\\
A = P.Q = \dfrac{{2x - 2}}{{\sqrt x }}.\dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\
= \dfrac{{2\left( {x - 1} \right)}}{{\sqrt x }}.\dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} = \dfrac{{2{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }}\\
c)A.\sqrt x < 8\\
\to \dfrac{{2{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }}.\sqrt x < 8\\
\to x - 2\sqrt x + 1 < 4\\
\to x - 2\sqrt x - 3 < 0\\
\to \left( {\sqrt x - 3} \right)\left( {\sqrt x + 1} \right) < 0\\
\to \sqrt x - 3 < 0\left( {do:\sqrt x + 1 > 0\forall x > 0} \right)\\
\to 0 < x < 9;x \ne 1
\end{array}\)
