Đáp án:
2)\(\dfrac{{76545}}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\,{\left( {x - \dfrac{2}{{\sqrt x }}} \right)^{12}} = \sum\limits_{k = 0}^{12} {C_{12}^k{x^{12 - k}}{{.2}^k}.\dfrac{1}{{{{\left( {\sqrt x } \right)}^k}}}} \\
= \sum\limits_{k = 0}^{12} {C_{12}^k.{x^{12 - \dfrac{{3k}}{2}}}{2^k}} \\
\Rightarrow 12 - \dfrac{{3k}}{2} = 1 \Leftrightarrow 24 - 3k = 2\\
\Leftrightarrow 22 = 3k \Leftrightarrow k = \dfrac{{22}}{3}\left( {ktm} \right)\\
2)\,{\left( {\dfrac{1}{{\sqrt 2 }} - 3{x^2}} \right)^{10}} = \sum\limits_{k = 0}^{10} {C_{10}^k{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^{10 - k}}.{{\left( { - 3} \right)}^k}{x^k}} \\
\Rightarrow k = 6 \Rightarrow he\,so:\,C_{10}^6.{\left( {\dfrac{1}{{\sqrt 2 }}} \right)^{10 - 6}}{\left( { - 3} \right)^6} = \dfrac{{76545}}{2}
\end{array}\)
