Thực hiện phép tính: $\frac{\sqrt[]{3} +\sqrt[]{2} - 1}{2 +\sqrt[]{6}}$ + $\frac{\sqrt[]{2}-\sqrt[]{3}}{\sqrt[]{2} +1}$ · ($\frac{\sqrt[]{3}}{2- \sqrt[]{6}}$ + $\frac{\sqrt[]{3}}{2+\sqrt[]{6}}$ )- $\frac{1}{\sqrt[]{2} }$ Giúp mình với ạ. Hứa sẽ vào cho mn
1 câu trả lời
Đáp án:
$\begin{array}{l}
\dfrac{{\sqrt 3 + \sqrt 2 - 1}}{{2 + \sqrt 6 }} + \dfrac{{\sqrt 2 - \sqrt 3 }}{{\sqrt 2 + 1}}.\left( {\dfrac{{\sqrt 3 }}{{2 - \sqrt 6 }} + \dfrac{{\sqrt 3 }}{{2 + \sqrt 6 }}} \right) - \dfrac{1}{{\sqrt 2 }}\\
= \dfrac{{\sqrt 3 + \sqrt 2 - 1}}{{\sqrt 2 \left( {\sqrt 2 + \sqrt 3 } \right)}}\\
+ \dfrac{{\sqrt 2 - \sqrt 3 }}{{\sqrt 2 + 1}}.\dfrac{{\sqrt 3 \left( {2 + \sqrt 6 } \right) + \sqrt 3 \left( {2 - \sqrt 6 } \right)}}{{\left( {2 - \sqrt 6 } \right)\left( {2 + \sqrt 6 } \right)}} - \dfrac{1}{{\sqrt 2 }}\\
= \dfrac{1}{{\sqrt 2 }} - \dfrac{1}{{2 + \sqrt 6 }}\\
+ \dfrac{{\sqrt 2 - \sqrt 3 }}{{\sqrt 2 + 1}}.\dfrac{{4\sqrt 3 }}{{\sqrt 2 .\left( {\sqrt 2 - \sqrt 3 } \right).\left( {2 + \sqrt 6 } \right)}} - \dfrac{1}{{\sqrt 2 }}\\
= - \dfrac{1}{{2 + \sqrt 6 }} + \dfrac{{2\sqrt 2 .\sqrt 3 }}{{\left( {\sqrt 2 + 1} \right).\left( {2 + \sqrt 6 } \right)}}\\
= \dfrac{{2\sqrt 6 - \left( {\sqrt 2 + 1} \right)}}{{\left( {\sqrt 2 + 1} \right)\left( {2 + \sqrt 6 } \right)}}\\
= \dfrac{{2\sqrt 6 - \sqrt 2 - 1}}{{\left( {\sqrt 2 + 1} \right)\left( {2 + \sqrt 6 } \right)}}\\
= \dfrac{{\left( {2\sqrt 6 - \sqrt 2 - 1} \right)\left( {\sqrt 2 - 1} \right)\left( {\sqrt 6 - 2} \right)}}{{\left( {2 - 1} \right).\left( {6 - {2^2}} \right)}}\\
= \dfrac{{\left( {4\sqrt 3 - 2\sqrt 6 - 2 + \sqrt 2 - \sqrt 2 + 1} \right)\left( {\sqrt 6 - 2} \right)}}{2}\\
= \dfrac{{\left( {4\sqrt 3 - 2\sqrt 6 - 1} \right)\left( {\sqrt 6 - 2} \right)}}{2}\\
= \dfrac{{12\sqrt 2 - 8\sqrt 3 - 12 + 4\sqrt 6 - \sqrt 6 + 2}}{2}\\
= \dfrac{{12\sqrt 2 - 8\sqrt 3 + 3\sqrt 6 - 10}}{2}
\end{array}$