Hỏi ai cũng bó tay !!! Mấy mod giúp em với ạ :( Rút gọn $\frac{6}{√2- √3 +3 }$
1 câu trả lời
Đáp án:
$\begin{array}{l}
\dfrac{6}{{\sqrt 2 - \sqrt 3 + 3}}\\
= \dfrac{{6\left( {\sqrt 2 - \sqrt 3 - 3} \right)}}{{\left( {\sqrt 2 - \sqrt 3 + 3} \right).\left( {\sqrt 2 - \sqrt 3 - 3} \right)}}\\
= \dfrac{{6.\left( {\sqrt 2 - \sqrt 3 - 3} \right)}}{{{{\left( {\sqrt 2 - \sqrt 3 } \right)}^2} - {3^2}}}\\
= \dfrac{{6.\left( {\sqrt 2 - \sqrt 3 - 3} \right)}}{{2 - 2\sqrt 6 + 3 - 9}}\\
= \dfrac{{6\left( {\sqrt 2 - \sqrt 3 - 3} \right)}}{{ - 4 - 2\sqrt 6 }}\\
= \dfrac{{3.\left( {3 + \sqrt 3 - \sqrt 2 } \right)}}{{2 + \sqrt 6 }}\\
= \dfrac{{3.\left( {3 + \sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 6 - 2} \right)}}{{\left( {2 + \sqrt 6 } \right)\left( {\sqrt 6 - 2} \right)}}\\
= \dfrac{{3\left( {3 + \sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 6 - 2} \right)}}{{6 - 4}}\\
= \dfrac{{\left( {9 + 3\sqrt 3 - 3\sqrt 2 } \right)\left( {\sqrt 6 - 2} \right)}}{2}\\
= \dfrac{{9\sqrt 6 - 18 + 3.3\sqrt 2 - 6\sqrt 3 - 3.2.\sqrt 3 + 6\sqrt 2 }}{2}\\
= \dfrac{{9\sqrt 6 + 15\sqrt 2 - 12\sqrt 3 - 18}}{2}
\end{array}$