Đáp án + Giải thích các bước giải:
Câu 7 : Đặt `S=1+2+2^2+2^3+...+2^2008` $\\$ `=> 2S=2+2^2+2^3+...+2^2009` $\\$ `=> 2S - S = (2+2^2+2^3+...+2^2009)-(1+2+2^2+2^3+...+2^2008)` $\\$ `=> S = 2^2009 - 1`
Do đó: `B = S/(1 - 2^2009)=(2^2009-1)/(1-2^2009)=(2^2009-1)/[-(2^2009-1)]=-1`
Vậy `B = -1`
Bài 1 ( đề 3):
`a)` `(-3)/4 + 3/7 + (-1)/4 + 4/9 + 4/7` $\\$ `= ((-3)/4+(-1)/4)+(3/7+4/7)+4/9=-1+1+4/9=4/9`
`b)` `(-7)/9*4/11 + (-7)/9*7/11+5 7/9` $\\$ `= (-7)/9(4/11+7/11) + 5 7/9` $\\$ `= (-7)/9 + 5 7/9 = -1+ 2/9 + 5 + 7/9` $\\$ `= (-1+5)+(2/9 + 7/9) = 4 + 1 = 5`
`c)` `(21/31 + 2013/6039)-(44/53-10/31)-9/53` $\\$ `= 21/31 + 2013/6039 - 44/53 + 10/31 - 9/53` $\\$ `= 21/31+1/3 - 44/53 + 10/31 - 9/53` $\\$ `= (21/31 + 10/31) + (-44/53 - 9/53) + 1/3` $\\$ `= 1 + (-1) + 1/3 = 1/3`
`d)` `6/7 + 5/8:5-3/16*(-2)^2` $\\$ `= 6/7+5/40-3/16*4` $\\$ `= 6/7+1/8-3/4*1` $\\$ `= 6/7 + 1/8 - 3/4= 13/56`
