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