Q=(15/(√6+1) +4/(√6-2) -12/(3-√6) . (√6+1) 12/07/2021 Bởi Josephine Q=(15/(√6+1) +4/(√6-2) -12/(3-√6) . (√6+1)
Giải thích các bước giải: Ta có: \(\begin{array}{l}Q = \left( {\dfrac{{15}}{{\sqrt 6 + 1}} + \dfrac{4}{{\sqrt 6 – 2}} – \dfrac{{12}}{{3 – \sqrt 6 }}} \right).\left( {\sqrt 6 + 1} \right)\\ = \left( {\dfrac{{15.\left( {\sqrt 6 – 1} \right)}}{{\left( {\sqrt 6 + 1} \right)\left( {\sqrt 6 – 1} \right)}} + \dfrac{{4.\left( {\sqrt 6 + 2} \right)}}{{\left( {\sqrt 6 – 2} \right)\left( {\sqrt 6 + 2} \right)}} – \dfrac{{12\left( {3 + \sqrt 6 } \right)}}{{\left( {3 – \sqrt 6 } \right)\left( {3 + \sqrt 6 } \right)}}} \right).\left( {\sqrt 6 + 1} \right)\\ = \left( {\dfrac{{15.\left( {\sqrt 6 – 1} \right)}}{{6 – 1}} + \dfrac{{4\left( {\sqrt 6 + 2} \right)}}{{6 – 4}} – \dfrac{{12.\left( {3 + \sqrt 6 } \right)}}{{9 – 6}}} \right).\left( {\sqrt 6 + 1} \right)\\ = \left[ {3.\left( {\sqrt 6 – 1} \right) + 2.\left( {\sqrt 6 + 2} \right) – 4.\left( {3 + \sqrt 6 } \right)} \right].\left( {\sqrt 6 + 1} \right)\\ = \left[ {3\sqrt 6 – 3 + 2\sqrt 6 + 4 – 12 – 4\sqrt 6 } \right].\left( {\sqrt 6 + 1} \right)\\ = \left( {\sqrt 6 – 11} \right)\left( {\sqrt 6 + 1} \right)\\ = 6 + \sqrt 6 – 11\sqrt 6 – 11\\ = – 5 – 10\sqrt 6 \end{array}\) Bình luận
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
Q = \left( {\dfrac{{15}}{{\sqrt 6 + 1}} + \dfrac{4}{{\sqrt 6 – 2}} – \dfrac{{12}}{{3 – \sqrt 6 }}} \right).\left( {\sqrt 6 + 1} \right)\\
= \left( {\dfrac{{15.\left( {\sqrt 6 – 1} \right)}}{{\left( {\sqrt 6 + 1} \right)\left( {\sqrt 6 – 1} \right)}} + \dfrac{{4.\left( {\sqrt 6 + 2} \right)}}{{\left( {\sqrt 6 – 2} \right)\left( {\sqrt 6 + 2} \right)}} – \dfrac{{12\left( {3 + \sqrt 6 } \right)}}{{\left( {3 – \sqrt 6 } \right)\left( {3 + \sqrt 6 } \right)}}} \right).\left( {\sqrt 6 + 1} \right)\\
= \left( {\dfrac{{15.\left( {\sqrt 6 – 1} \right)}}{{6 – 1}} + \dfrac{{4\left( {\sqrt 6 + 2} \right)}}{{6 – 4}} – \dfrac{{12.\left( {3 + \sqrt 6 } \right)}}{{9 – 6}}} \right).\left( {\sqrt 6 + 1} \right)\\
= \left[ {3.\left( {\sqrt 6 – 1} \right) + 2.\left( {\sqrt 6 + 2} \right) – 4.\left( {3 + \sqrt 6 } \right)} \right].\left( {\sqrt 6 + 1} \right)\\
= \left[ {3\sqrt 6 – 3 + 2\sqrt 6 + 4 – 12 – 4\sqrt 6 } \right].\left( {\sqrt 6 + 1} \right)\\
= \left( {\sqrt 6 – 11} \right)\left( {\sqrt 6 + 1} \right)\\
= 6 + \sqrt 6 – 11\sqrt 6 – 11\\
= – 5 – 10\sqrt 6
\end{array}\)