k trước thì mk giải cho

c)(1-3+5-7+....+17)+(-2+4-6+....-18)

gọiA = (1-3+5-7+ ....+17)

=(1-3)+(5-7)+....+(13-15)+17

=(-2)+(-2)+....(-2)+17

Có tất cả số số hạng từ 15 đến 1 là:

       (15-1):2+1=8 (số)

    tổng trên là:

        [(15+1).8:2]+17=81

  còn phần (-2+4-6+....-18) làm tương tự nha

Vương Nguyên k nha

ta có: B= -1+4-5+8-9+12-...-53+56

B= -1+-1+-1+-1+...+-1+56

B= -1.( 1+50:2)+56

B= -1.26+56

B= -26+56

B= 30

vậy b= 30

ta có:C=(1-2-3+4)+(5-6-7+8)+...+(97-98-99+100)

C= 0+0+...+0

C= 0

vậy C=0

c: C=1*2+2*3+3*4+...+58*59+59*60

=>3*C=1*2*3+2*3*(4-1)+3*4*(5-2)+...+58*59(60-57)+59*60(61-58)

=>3*C=1*2*3+2*3*4-1*2*3+...+58*59*60-58*59*57+59*60*61-58*59*60

=>3*C=59*60*61

=>C=59*20*61=71980

a,  15 . { 32 : [ 6 - 5 + 5 ( 9 : 3 ) ] + 3 } - 2018 0

= 15.{32:[1+15]+3}–1

= 15.5–1

= 74

b,  25 . { 2 7 : [ 12 - 4 + 2 2 . 16 : 2 3 ] - 2 4 }

= 25.{128:[8+4.2]–16}

= 25.24

= 600

c,  2019 . { 101 - 1000 : [ 2 2 . 2 3 + 5 6 : 5 3 - 6 2 : 11 - 2018 0 ] }

= 2019.{101–1000:[(32+125–36):11–1]}

= 2019.{101–1000:[121:11–1]}

= 2019.{101–1000:10}

= 2019.1

= 2019

1. 

$(5^{1986}-5^{1985}):5^{1985}=5^{1985}(5-1):5^{1985}=5-1=4$

2.

\((7^{846}+7^{847}):7^{846}=7^{846}(1+7):7^{846}=1+7=8\)

3.

\((9^{2018}-3^{4036}):6^{2006}=[(3^2)^{2018}-3^{4036}]:6^{2006}\)

$=(3^{4036}-3^{4036}):6^{2006}=0:6^{2006}=0$

4.

$(7^{80}.8^{70}-56^{70}):56^{70}$

$=[7^{10}(7.8)^{70}-56^{70}]:56^{70}$

$=[7^{10}.56^{70}-56^{70}]:56^{70}$

$=56^{70}(7^{10}-1):56^{70}=7^{10}-1$

5.

$4^{4016}:(4^{4017}-4^{4016})=4^{4016}:[4^{4016}(4-1)]$

$=4^{4016}:4^{4016}:3=1:3=\frac{1}{3}$

6.

$(12^{206}.2^{207}-24^{206}):24^{206}$

$=(12^{206}.2^{206}.2-24^{206}):24^{206}$

$=[(12.2)^{206}.2-24^{206}]:24^{206}$

$=(24^{206}.2-24^{206}):24^{206}$

$=24^{206}(2-1):24^{206}=2-1=1$

7.

$(5^2-24)^{8946}+4^{30}:2^{60}=1^{8946}+(2^2)^{30}:2^{60}$

$=1+2^{60}:2^{60}=1+1=2$

8.

$(37.8^{1007}-7.2^{3021}):8^{1007}=[37.8^{1007}-7.(2^3)^{1007}]:8^{1007}$

$=[37.8^{1007}-7.8^{1007}]:8^{1007}$

$=8^{1007}(37-7):8^{1007}=37-7=30$