Đáp án:
c. x=9
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne 1\\
A = \left[ {\dfrac{{x + 2\sqrt x - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}} \right]:\dfrac{{\sqrt x + 2}}{{x + \sqrt x + 1}}\\
= \dfrac{{\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{x + \sqrt x + 1}}{{\sqrt x + 2}}\\
= \dfrac{1}{{\sqrt x + 2}}\\
b.Thay:x = 9 - 4\sqrt 5 \\
= 5 - 2.2.\sqrt 5 + 4 = {\left( {\sqrt 5 - 2} \right)^2}\\
\to A = \dfrac{1}{{\sqrt {{{\left( {\sqrt 5 - 2} \right)}^2}} + 2}}\\
= \dfrac{1}{{\sqrt 5 - 2 + 2}} = \dfrac{1}{{\sqrt 5 }}\\
c.A = \dfrac{1}{5}\\
\to \dfrac{1}{{\sqrt x + 2}} = \dfrac{1}{5}\\
\to \sqrt x + 2 = 5\\
\to \sqrt x = 3\\
\to x = 9\left( {TM} \right)
\end{array}\)
