Đáp án:
\[2\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
P = \frac{{4 + \sqrt 7 }}{{3\sqrt 2 + \sqrt {4 + \sqrt 7 } }} + \frac{{4 - \sqrt 7 }}{{3\sqrt 2 - \sqrt {4 - \sqrt 7 } }}\\
= \frac{{8 + 2\sqrt 7 }}{{6 + \sqrt {8 + 2\sqrt 7 } }} + \frac{{8 - 2\sqrt 7 }}{{6 - \sqrt {8 - 2\sqrt 7 } }}\\
= \frac{{7 + 2\sqrt 7 + 1}}{{6 + \sqrt {7 + 2\sqrt 7 + 1} }} + \frac{{7 - 2\sqrt 7 + 1}}{{6 - \sqrt {7 - 2\sqrt 7 + 1} }}\\
= \frac{{{{\left( {\sqrt 7 + 1} \right)}^2}}}{{6 + \sqrt {{{\left( {\sqrt 7 + 1} \right)}^2}} }} + \frac{{{{\left( {\sqrt 7 - 1} \right)}^2}}}{{6 - \sqrt {{{\left( {\sqrt 7 - 1} \right)}^2}} }}\\
= \frac{{{{\left( {\sqrt 7 + 1} \right)}^2}}}{{6 + \sqrt 7 + 1}} + \frac{{{{\left( {\sqrt 7 - 1} \right)}^2}}}{{6 - \sqrt 7 + 1}}\\
= \frac{{{{\left( {\sqrt 7 + 1} \right)}^2}}}{{7 + \sqrt 7 }} + \frac{{{{\left( {\sqrt 7 - 1} \right)}^2}}}{{7 - \sqrt 7 }}\\
= \frac{{{{\left( {\sqrt 7 + 1} \right)}^2}}}{{\sqrt 7 \left( {\sqrt 7 + 1} \right)}} + \frac{{{{\left( {\sqrt 7 - 1} \right)}^2}}}{{\sqrt 7 \left( {\sqrt 7 - 1} \right)}}\\
= \frac{{\sqrt 7 + 1}}{{\sqrt 7 }} + \frac{{\sqrt 7 - 1}}{{\sqrt 7 }} = \frac{{2\sqrt 7 }}{{\sqrt 7 }} = 2
\end{array}\)
