tìm hệ số của x^5 trong khai triển P=x(1-2x)^5 + x^2(1+3x)^10

<p><p style="text-align: left;">Đ&aacute;p &aacute;n:</p></p><p><p style="text-align: left;">\(6600\)</p></p><p><p style="text-align: left;">Giải th&iacute;ch c&aacute;c bước giải:</p></p><p><p style="text-align: left;">\(\begin{array}{l}<br />x{\left( {1 - 2x} \right)^5} = x\sum\limits_{k = 0}^5 {C_5^k.{{\left( { - 2} \right)}^k}{x^k}} &nbsp;= \sum\limits_{k = 0}^5 {C_5^k.{{\left( { - 2} \right)}^k}{x^{k + 1}}} \<br />&nbsp; \cr \Rightarrow k + 1 = 5 \cr \Rightarrow k = 4\<br />&nbsp; \cr \Rightarrow \,he\,so\,cua\,{x^5}\,la\,\,C_5^4.{\left( { - 2} \right)^4}\<br />{x^2}{\left( {1 + 3x} \right)^{10}} = {x^2}\sum\limits_{{k_1} = 0}^{10} {C_{10}^{{k_1}}{{.3}^{{k_1}}}{x^{{k_1}}}} &nbsp;= \sum\limits_{{k_1} = 0}^{10} {C_{10}^{{k_1}}{{.3}^{{k_1}}}{x^{{k_1} + 2}}} \<br />&nbsp; \cr \Rightarrow {k_1} + 2 = 5 \cr \Rightarrow {k_1} = 3\<br />&nbsp; \cr \Rightarrow \,he\,so\,cua\,{x^5}\,la\,\,C_{10}^3{.3^3}\<br />&nbsp; \cr \Rightarrow \,he\,so\,cua\,{x^5}\,trong\,P\,la\,\,C_5^4.{\left( { - 2} \right)^4} + C_{10}^3{.3^3} = 6600<br />\end{array}\)</p></p>