Đáp án:
c. \(x = 3 + 2\sqrt 2 \)
Giải thích các bước giải:
\(\begin{array}{l}
a.M = \left[ {\dfrac{{\sqrt x + x + \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}} \right]:\dfrac{{\sqrt x + 1}}{{x + \sqrt x + 1}}\\
= \left[ {\dfrac{{x + 2\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}} \right].\dfrac{{x + \sqrt x + 1}}{{\sqrt x + 1}}\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{x + \sqrt x + 1}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
b.Thay:x = \sqrt {3 + 2\sqrt 2 } - \sqrt {3 - 2\sqrt 2 } \\
= \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} - \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} \\
= \sqrt 2 + 1 - \sqrt 2 + 1 = 2\\
M = \dfrac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}} = \dfrac{{2 + 2\sqrt 2 + 1}}{{2 - 1}} = 3 + 2\sqrt 2 \\
c.M = \sqrt x \\
\to \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} = \sqrt x \\
\to \sqrt x + 1 = x - \sqrt x \\
\to x - 2\sqrt x - 1 = 0\\
\to \left[ \begin{array}{l}
\sqrt x = 1 + \sqrt 2 \\
\sqrt x = 1 - \sqrt 2 \left( l \right)
\end{array} \right.\\
\to x = 3 + 2\sqrt 2
\end{array}\)
