Đáp án:
c) \(B = \sqrt 2 \)
Giải thích các bước giải:
\(\begin{array}{l}
a)B = \left[ {\dfrac{{2x + 1 - \sqrt x \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}} \right].\left[ {\dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\sqrt x + 1}} - \sqrt x } \right]\\
= \dfrac{{2x + 1 - x + \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\left( {x - 2\sqrt x + 1} \right)\\
= \dfrac{{x + \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.{\left( {\sqrt x - 1} \right)^2}\\
= \sqrt x - 1\\
b)B = \dfrac{1}{3}\\
\to \sqrt x - 1 = \dfrac{1}{3}\\
\to \sqrt x = \dfrac{4}{3}\\
\to x = \dfrac{{16}}{9}\\
c)Thay:x = 3 + 2\sqrt 2 = 2 + 2\sqrt 2 + 1\\
= {\left( {\sqrt 2 + 1} \right)^2}\\
\to B = \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} - 1 = \sqrt 2 + 1 - 1 = \sqrt 2
\end{array}\)
