Đáp án:
36B
37B
Giải thích các bước giải:
\(\eqalign{
& Cau\,\,36: \cr
& {\left( {2x - {1 \over {{x^2}}}} \right)^6} = {\left( {2x - {x^{ - 2}}} \right)^6} \cr
& = \sum\limits_{k = 0}^6 {C_6^k{2^k}{x^k}{{\left( { - 1} \right)}^{6 - k}}{{\left( {{x^{ - 2}}} \right)}^{6 - k}}} \cr
& = \sum\limits_{k = 0}^6 {C_6^k{2^k}{{\left( { - 1} \right)}^{6 - k}}{x^{ - 12 + 3k}}} \cr
& So\,\,hang\,\,khong\,\,chua\,\,x\,\,ung\,\,voi\,\, \cr
& - 12 + 3k = 0 \Leftrightarrow k = 4\,\,\left( {tm} \right) \cr
& Vay\,\,so\,\,hang\,\,khong\,\,chua\,\,x\,\,la: \cr
& C_6^4{2^4}{\left( { - 1} \right)^2} = 240. \cr
& Chon\,\,B. \cr
& Cau\,\,37: \cr
& {\left( {{1 \over x} + {x^3}} \right)^{10}} = {\left( {{x^{ - 1}} + {x^3}} \right)^{10}} \cr
& = \sum\limits_{k = 0}^{10} {C_{10}^k{{\left( {{x^{ - 1}}} \right)}^{10 - k}}{{\left( {{x^3}} \right)}^k}} \cr
& = \sum\limits_{k = 0}^{10} {C_{10}^k{x^{ - 10 + k + 3k}}} = \sum\limits_{k = 0}^{10} {C_{10}^k{x^{ - 10 + 4k}}} \cr
& So\,\,hang\,\,\,chua\,\,{x^6}\,\,ung\,\,voi\,\, - 10 + 4k = 6 \cr
& \Leftrightarrow 4k = 16 \Leftrightarrow k = 4\,\,\left( {tm} \right) \cr
& Vay\,he\,\,so\,\,cua\,\,{x^6}\,\,la:\,\,C_{10}^6 = 210. \cr
& Chon\,\,B. \cr} \)
