Đáp án:
c. \(0 \le x < 25\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne 25\\
A = \dfrac{{\sqrt x \left( {\sqrt x + 5} \right) - 10\sqrt x - 5\left( {\sqrt x - 5} \right)}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}\\
= \dfrac{{x + 5\sqrt x - 10\sqrt x - 5\sqrt x + 25}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}\\
= \dfrac{{x - 10\sqrt x + 25}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}\\
= \dfrac{{{{\left( {\sqrt x - 5} \right)}^2}}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}\\
= \dfrac{{\sqrt x - 5}}{{\sqrt x + 5}}\\
b.Thay:x = 6 - 2\sqrt 5 \\
= 5 - 2.\sqrt 5 .1 + 1 = {\left( {\sqrt 5 - 1} \right)^2}\\
\to A = \dfrac{{\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} - 5}}{{\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} + 5}} = \dfrac{{\sqrt 5 - 1 - 5}}{{\sqrt 5 - 1 + 5}}\\
= \dfrac{{\sqrt 5 - 6}}{{\sqrt 5 + 4}}\\
c.A < 0 \to \dfrac{{\sqrt x - 5}}{{\sqrt x + 5}} < 0\\
\to \sqrt x - 5 < 0\left( {do:\sqrt x + 5 > 0\forall x \ge 0} \right)\\
\to \sqrt x < 5\\
\to x < 25\\
\to 0 \le x < 25
\end{array}\)
