Toán Lớp 9: Thực hiện phép tính:
$\frac{\sqrt[]{3} +\sqrt[]{2} – 1}{2 +\sqrt[]{6}}$ + $\frac{\sqrt[]{2}-\sqrt[]{3}}{\sqrt[]{2} +1}$ · ($\frac{\sqrt[]{3}}{2- \sqrt[]{6}}$ + $\frac{\sqrt[]{3}}{2+\sqrt[]{6}}$ )- $\frac{1}{\sqrt[]{2} }$
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\dfrac{{\sqrt 3 + \sqrt 2 – 1}}{{2 + \sqrt 6 }} + \dfrac{{\sqrt 2 – \sqrt 3 }}{{\sqrt 2 + 1}}.\left( {\dfrac{{\sqrt 3 }}{{2 – \sqrt 6 }} + \dfrac{{\sqrt 3 }}{{2 + \sqrt 6 }}} \right) – \dfrac{1}{{\sqrt 2 }}\\
= \dfrac{{\sqrt 3 + \sqrt 2 – 1}}{{\sqrt 2 \left( {\sqrt 2 + \sqrt 3 } \right)}}\\
+ \dfrac{{\sqrt 2 – \sqrt 3 }}{{\sqrt 2 + 1}}.\dfrac{{\sqrt 3 \left( {2 + \sqrt 6 } \right) + \sqrt 3 \left( {2 – \sqrt 6 } \right)}}{{\left( {2 – \sqrt 6 } \right)\left( {2 + \sqrt 6 } \right)}} – \dfrac{1}{{\sqrt 2 }}\\
= \dfrac{1}{{\sqrt 2 }} – \dfrac{1}{{2 + \sqrt 6 }}\\
+ \dfrac{{\sqrt 2 – \sqrt 3 }}{{\sqrt 2 + 1}}.\dfrac{{4\sqrt 3 }}{{\sqrt 2 .\left( {\sqrt 2 – \sqrt 3 } \right).\left( {2 + \sqrt 6 } \right)}} – \dfrac{1}{{\sqrt 2 }}\\
= – \dfrac{1}{{2 + \sqrt 6 }} + \dfrac{{2\sqrt 2 .\sqrt 3 }}{{\left( {\sqrt 2 + 1} \right).\left( {2 + \sqrt 6 } \right)}}\\
= \dfrac{{2\sqrt 6 – \left( {\sqrt 2 + 1} \right)}}{{\left( {\sqrt 2 + 1} \right)\left( {2 + \sqrt 6 } \right)}}\\
= \dfrac{{2\sqrt 6 – \sqrt 2 – 1}}{{\left( {\sqrt 2 + 1} \right)\left( {2 + \sqrt 6 } \right)}}\\
= \dfrac{{\left( {2\sqrt 6 – \sqrt 2 – 1} \right)\left( {\sqrt 2 – 1} \right)\left( {\sqrt 6 – 2} \right)}}{{\left( {2 – 1} \right).\left( {6 – {2^2}} \right)}}\\
= \dfrac{{\left( {4\sqrt 3 – 2\sqrt 6 – 2 + \sqrt 2 – \sqrt 2 + 1} \right)\left( {\sqrt 6 – 2} \right)}}{2}\\
= \dfrac{{\left( {4\sqrt 3 – 2\sqrt 6 – 1} \right)\left( {\sqrt 6 – 2} \right)}}{2}\\
= \dfrac{{12\sqrt 2 – 8\sqrt 3 – 12 + 4\sqrt 6 – \sqrt 6 + 2}}{2}\\
= \dfrac{{12\sqrt 2 – 8\sqrt 3 + 3\sqrt 6 – 10}}{2}
\end{array}$