Toán Lớp 9: Hỏi ai cũng bó tay !!! Mấy mod giúp em với ạ 🙁
Rút gọn $\frac{6}{√2- √3 +3 }$
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Toán Lớp 9: Hỏi ai cũng bó tay !!! Mấy mod giúp em với ạ 🙁
Rút gọn $\frac{6}{√2- √3 +3 }$
Comments ( 1 )
\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}$