\(A=\left(1+3+5+...+2017\right)-\left(2+4+6+...+2018\right)\)

Tính tổng của hai cấp số cộng tự làm nốt nhé

a) 10-11+12-13+14-15+...-2017+2018

=(10-11)+(12-13)+(14-15)+...+(2016-2017)+2018

=(-1)+(-1)+(-1)+...+(-1)+2018

=(-1).1004+2018

=-1004+2018

=1004

b) -5+6-7+8-9+10-...-2017+2018

=-5+(6-7)+(8-9)+...+(2016-2017)+2018

=(-5)+(-1)+(-1)+...+(-1)+2018

=(-5)+(-1).1006+2018

=(-5)+(-1006)+2018

=1007

c) 3-4+5-6+7-8+...+101-102+103

= (3-4)+(5-6)+(7-8)+...+(101-102)+103

= (-1)+(-1)+(-1)+...+(-1)+103

= (-1).50+103 = (-50)+103

= 53

Câu a) phải là 1014

A = ( 2016 + 2017 ) - ( 2017 + 2018 ) + ( 2018 - 16 )

A = 2016 + 2017 - 2017 - 2018 + 2018 - 16

A = ( 2016 - 16 ) + ( 2017 - 2017 ) + ( 2018 - 2018 )

A = 2000 + 0 + 0

A = 2000

B = ( 157 - 215 ) + ( 315 - 157 )  + ( 215 - 265 )

B = 157 - 215 + 315 - 157 + 215 - 265

B = ( 157 - 157 ) + ( 215 - 215 ) + ( 315 - 265 )

B = 0 + 0 + 50

B = 50

gấp lắm rồi :<<

cứu mình :<<<

a) Các số có dạng : \(\frac{1}{a\left(a+1\right)}=\frac{\left(a+1\right)-a}{a\left(a+1\right)}=\frac{1}{a}-\)\(\frac{1}{a+1}\)

Thế vào bởi các số sẽ có kết quả

b) Các số có dạng : \(\frac{1}{a\left(a+2\right)}=\frac{1}{2}.\frac{2}{a\left(a+2\right)}=\frac{1}{2}.\frac{\left(a+2\right)-a}{a\left(a+2\right)}\)\(=\frac{1}{2}.\left(\frac{1}{a}-\frac{1}{a+2}\right)\)

Làm tương tự trên

c) Lấy nhân tử chung là 5 rồi làm như câu a)

bạn có thể làm ra hộ mình được ko mình ko hiểu

a: \(75\%+1,2-2+\frac15+2018^0\)

=0,75+1,2-2+0,2+1

=0,75+1,4-1

=0,75+0,4

=1,15

b: \(\left(-\frac43+0,75\right):\frac{2017}{2018}+\left(1\frac13-75\%\right):\frac{2017}{2018}\)

\(=\left(-\frac43+0,75+\frac43-0,75\right):\frac{2017}{2018}\)

=0

c: \(\left(2018-\frac13-\frac24-\cdots-\frac{2018}{2020}\right):\left(\frac{1}{15}+\frac{1}{20}+\cdots+\frac{1}{10100}\right)\)

\(=\left(1-\frac13+1-\frac24+\cdots+1-\frac{2018}{2020}\right):\left\lbrack\frac15\left(\frac13+\frac14+\cdots+\frac{1}{2020}\right)\right\rbrack\)

\(=\frac{\left(\frac23+\frac24+\cdots+\frac{2}{2020}\right)}{\frac15\left\lbrack\frac13+\frac14+\cdots+\frac{1}{2020}\right\rbrack}=2:\frac15=10\)

Giải:

a) \(75\%+1,2-2+\dfrac{1}{5}+2018^0\) 

=\(\dfrac{3}{4}+\dfrac{6}{5}-2+\dfrac{1}{5}+1\) 

=\(\left(\dfrac{6}{5}+\dfrac{1}{5}\right)+\left(\dfrac{3}{4}-2+1\right)\) 

=\(\dfrac{7}{5}+\dfrac{-1}{4}\) 

=\(\dfrac{23}{20}\) 

b) \(\left(\dfrac{-4}{3}+0,75\right):\dfrac{2017}{2018}+\left(1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\) 

=\(\left(\dfrac{-4}{3}+0,75+1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\) 

=\(\left[\left(\dfrac{-4}{3}+1+\dfrac{1}{3}\right)+\left(0,75-75\%\right)\right]:\dfrac{2017}{2018}\) 

=\(\left[0+0\right]:\dfrac{2017}{2018}\) 

=0\(:\dfrac{2017}{2018}\) 

=0

c)\(\left(2018-\dfrac{1}{3}-\dfrac{2}{4}-\dfrac{3}{5}-\dfrac{4}{6}-...-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\)

=\(\left(1-\dfrac{1}{3}-1-\dfrac{2}{4}-1-\dfrac{3}{5}-1-\dfrac{4}{6}-...-1-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) 

=\(\left(\dfrac{2}{3}-\dfrac{2}{4}-\dfrac{2}{5}-\dfrac{2}{6}-...-\dfrac{2}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left[2.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[\dfrac{5}{5}.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[5.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(10.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =-10