a)-1-2-3-4-5-6-....-80

=(-1)+(-2)+(-3)+(-4)+(-5)+(-6)+...+(-80)

Khoảng cách giữa các số:(-1)-(-2)=1

Tổng các số hạng:(-1)-(-80)+1=80 số

Tổng:[(-1)+(-80)].80:2= -3240

=>-1-2-3-4-5-6+......-80=-3240

b,1-2+3-4+5-6+......+2021-2022

=(1-2)+(3-4)+(5-6)+...+(2021-2022)

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

Tổng số cặp là:

(2022-1+1):2=1011 cặp

-1.1011=-1011

=>1-2+3-4+5-6+......+2021-2022= -1011

c, Đề bài sai

d,-4-8-12-16-.......-2020

=-4+(-8)+(-12)+(-16)+...+(-2020)

Khoảng cách giữa các số:-4-(-8)=4

Tổng các số hạng:[-4-(-2020]:4+1=505 số

Tổng:[-4+(-2020)].505:2=-511060

=>-4-8-12-16-.......-2020=-511060

Mọi người giúp mình với ạ! Thank you everyone!

Ghi đầy đủ nha

bn có thể ghi rõ ràng đc ko?

a)\(-\dfrac{3}{7}+\dfrac{5}{13}+\dfrac{3}{7}\) 

=\(\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\dfrac{5}{13}\) 

=\(0+\dfrac{5}{13}\) 

=\(\dfrac{5}{13}\)

A=1+2+3+4+5+...+50
A=(50+1)+(49+2)+(48+3)+...
A=(50+1)*[(50-1):1+1]:2
A=51*25=1275
B=2+4+6+8+10+...+100
B=(100+2)+(98+4)+(96+6)+...
B=(100+2)*[(100-2):2+1]:2
B=102*25=2550
C=1+4+7+10+13+...+99
C=(99+1)+(96+4)+(93+7)+...
C=(99+1)*[(99-1):3+1]:2
C=100*16.8333=1683.33
D=2+5+8+11+14+...+98
D=(98+2)+(95+5)+(92+8)+...
D=(98+2)*[(98-2):3+1]:2
D=100*16.5=1650
E=1+2+3+4+5+...+25
E=(25+1)+(24+2)+(23+3)+...
E=(25+1)*[(25-1):1+1]:2
E=26*12.5=325
F=2+4+6+8+10+...+50
F=(50+2)+(48+4)+(46+6)+...
F=(50+2)*[(50-2):2+1]:2
F=52*12.5=650
G=3+5+7+9+11+...+51
G=(51+3)+(49+5)+(47+7)+...
G=(51+3)*[(51-3):2+1]:2
G=54*12.5=675
H=1+5+9+13+17+...+81
H=(81+1)+(77+5)+(73+9)+...
H=(81+1)*[(81-1):4+1]:2
H=82*10.5=861

a) A =1 + 2 + 3 + 4 + … + 50

Số số hạng của dãy số trên là:

(50 - 1) : 1 + 1 = 50 (số số hạng)

  A =(1+ 50) . 50 : 2

      = 51 . 50 : 2

      = 2550 : 2

      = 1275

b) B = 2 + 4 + 6 + 8 + ... + 100

Số số hạng của dãy số trên là:

(100 - 2) : 2 + 1 = 50 (số hạng)

Có số cặp là:

50 : 2 = 25 (cặp)

Tổng của 1 cặp là:

100 + 2 = 102

Tổng của dãy số là:

25 .102 = 2550

c) C = 1 + 3 + 5 + 7 + … + 99

Số số hạng của dãy trên là:

(99 - 1) : 2 + 1 = 50 (số số hạng)

C = (1 + 99) . 50 : 2

  = 100 . 50 : 2

  = 5000 : 2

  = 2500

d) D = 2 + 5 + 8 + 11 + … + 98

Số số hạng của dãy trên là:

 (98 - 2) : 3 + 1 = 33 (số số hạng)

=> Dãy trên có 16 cặp

D = (95 + 2) .16 + 98

   = 97 . 16 + 98

   = 1552 +98

   = 1650

1: =72/90+65/90=137/90

2: =24/56-77/56=-53/56

3: =-7/10+4/5=1/10

4: =15/100-4/100=11/100

5: =4/6-5/6=-1/6

6: =10/40-15/40-76/40=-81/40

7: =-9/10+7/18

=-81/90+35/90=-46/90=-23/45

8: =27/90-55/90=-28/90=-14/45

9: =36/60-50/60-35/60=-49/60

10: =-4/9+5/6-3/8

=-32/72+60/72-27/72

=1/72

\(1,\dfrac{4}{5}+\dfrac{13}{18}=\dfrac{72}{90}+\dfrac{65}{90}=\dfrac{137}{90}\)

\(2,\dfrac{3}{7}-\dfrac{11}{8}=\dfrac{24}{56}-\dfrac{77}{56}=\dfrac{-53}{56}\)

\(3,-\dfrac{7}{10}-\left(-\dfrac{4}{5}\right)=-\dfrac{7}{10}-\left(-\dfrac{8}{10}\right)=\dfrac{1}{10}\)

\(4,\dfrac{3}{20}-\dfrac{1}{25}=\dfrac{75}{500}-\dfrac{20}{500}=\dfrac{55}{500}=\dfrac{11}{100}\)

\(5,\dfrac{2}{3}-\dfrac{5}{6}=\dfrac{12}{18}-\dfrac{15}{18}=-\dfrac{3}{18}=-\dfrac{1}{6}\)

\(6,\dfrac{1}{4}+\left(-\dfrac{3}{8}\right)-\dfrac{19}{10}=\dfrac{8}{32}+\left(-\dfrac{12}{32}\right)-\dfrac{19}{10}=-\dfrac{4}{32}-\dfrac{19}{10}\)

\(=-\dfrac{1}{8}-\dfrac{19}{10}=-\dfrac{10}{80}-\dfrac{152}{80}=-\dfrac{162}{80}=-\dfrac{81}{40}\)

\(7,-\dfrac{9}{10}-\left(-\dfrac{7}{18}\right)=-\dfrac{162}{180}-\left(-\dfrac{70}{180}\right)=-\dfrac{92}{180}=-\dfrac{23}{45}\)

\(8,\dfrac{3}{10}-\dfrac{11}{18}=\dfrac{54}{180}-\dfrac{110}{180}=-\dfrac{56}{180}=-\dfrac{14}{45}\)

\(9,\dfrac{3}{5}-\dfrac{5}{6}+\left(-\dfrac{7}{12}\right)=\dfrac{18}{30}-\dfrac{25}{30}+\left(-\dfrac{7}{12}\right)=-\dfrac{7}{30}+\left(-\dfrac{7}{12}\right)\)

\(=-\dfrac{84}{360}+\left(-\dfrac{210}{360}\right)=-\dfrac{294}{360}=-\dfrac{49}{60}\)

\(10,-\dfrac{4}{9}-\dfrac{-5}{6}-\dfrac{3}{8}=-\dfrac{24}{54}-\dfrac{-45}{54}-\dfrac{3}{8}\)

\(=\dfrac{21}{54}-\dfrac{3}{8}=\dfrac{7}{18}-\dfrac{3}{8}=\dfrac{56}{144}-\dfrac{54}{144}=\dfrac{2}{144}=\dfrac{1}{72}\)

`@mt`

2 câu c,d làm tương tự  

5:

a: \(3^{2n}=\left(3^2\right)^n=9^n\)

\(\left(2^{3n}\right)=\left(2^3\right)^n=8^n\)

=>\(3^{2n}>2^{3n}\)

b: \(199^{20}=\left(199^4\right)^5=1568239201^5\)

\(2003^{15}=\left(2003^3\right)^5=8036054027^5\)

mà \(1568239201< 8036054027\)

nên \(199^{20}< 2003^{15}\)

4: \(100< 5^{2x-1}< 5^6\)

mà \(25< 100< 125\)

nên \(125< 5^{2x-1}< 5^6\)

=>3<2x-1<6

=>4<2x<7

=>2<x<7/2

mà x nguyên

nên x=3

a) 17.13+17.42-17.35

=17.(13+42-35)

=17.20=340

b) [25.(18-4²)-10]:4+6

=(25.2-10):4+6

=40:4+6=16

c) 3⁶:3²+2³.2²-3².3

=3⁴+2⁵-3³

=81+32-27=86

d) B=3.4²-2².3

=3.(16-4)

=3.12=36

e)2022⁰+3.[5².10-(23-13)²]

=1+3.(250-100)

=1+450=451

g) 27.77+24.27-27

=27.(77+24-1)

=27.100=2700

h) 5.2³+7⁹:7⁷-1²⁰²⁰

=40+7²-1

=89-1=88

i) 120:{54[50:2+(3²-2.4)]}

=120:[54(25+1)]

=120:1404=10/117

a: \(\left(-\dfrac{5}{6}+\dfrac{2}{5}\right):\dfrac{3}{8}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right):\dfrac{3}{8}\)

\(=\left(-\dfrac{5}{6}+\dfrac{2}{5}\right)\cdot\dfrac{8}{3}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right)\cdot\dfrac{8}{3}\)

\(=\dfrac{8}{3}\left(-\dfrac{5}{6}+\dfrac{2}{5}+\dfrac{4}{5}-\dfrac{11}{30}\right)\)

\(=\dfrac{8}{3}\cdot\dfrac{-25+36-11}{30}\)

=0

b: \(\left(-\dfrac{3}{4}+\dfrac{2}{5}\right):\dfrac{3}{7}+\left(\dfrac{3}{5}+\dfrac{-1}{4}\right):\dfrac{3}{7}\)

\(=\left(-\dfrac{3}{4}+\dfrac{2}{5}\right)\cdot\dfrac{7}{3}+\left(\dfrac{3}{5}-\dfrac{1}{4}\right)\cdot\dfrac{7}{3}\)

\(=\dfrac{7}{3}\left(-\dfrac{3}{4}+\dfrac{2}{5}+\dfrac{3}{5}-\dfrac{1}{4}\right)\)

\(=\dfrac{7}{3}\cdot0=0\)

c: \(\dfrac{-13}{18}\cdot\dfrac{5}{8}+\dfrac{-5}{18}\cdot\dfrac{2}{9}+\dfrac{-13}{18}\cdot\dfrac{3}{8}+\dfrac{-5}{18}\cdot\dfrac{7}{9}\)

\(=\left(-\dfrac{13}{18}\cdot\dfrac{5}{8}+\dfrac{-13}{18}\cdot\dfrac{3}{8}\right)+\left(-\dfrac{5}{18}\cdot\dfrac{2}{9}+\dfrac{-5}{18}\cdot\dfrac{7}{9}\right)\)

\(=-\dfrac{13}{18}\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{-5}{18}\left(\dfrac{2}{9}+\dfrac{7}{9}\right)\)

\(=-\dfrac{13}{18}-\dfrac{5}{18}=-\dfrac{18}{18}=-1\)

d: Sửa đề: \(\dfrac{-11}{19}\cdot\dfrac{4}{9}+\dfrac{-8}{19}\cdot\dfrac{3}{7}+\dfrac{-11}{19}\cdot\dfrac{5}{9}+\dfrac{-8}{19}\cdot\dfrac{4}{7}\)

\(=\left(-\dfrac{11}{19}\cdot\dfrac{4}{9}+\dfrac{-11}{19}\cdot\dfrac{5}{9}\right)+\left(\dfrac{-8}{19}\cdot\dfrac{3}{7}+\dfrac{-8}{19}\cdot\dfrac{4}{7}\right)\)

\(=-\dfrac{11}{19}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)+\dfrac{-8}{19}\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\)

\(=-\dfrac{11}{19}-\dfrac{8}{19}=-\dfrac{19}{19}=-1\)

\(a.\left(-\dfrac{5}{6}+\dfrac{2}{5}\right):\dfrac{3}{8}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right):\dfrac{3}{8}\)

\(=\left(-\dfrac{13}{30}\right):\dfrac{3}{8}+\dfrac{13}{30}:\dfrac{3}{8}\)

\(=\left[\left(-\dfrac{13}{30}+\dfrac{13}{30}\right)\right]:\dfrac{3}{8}\)

\(=0:\dfrac{3}{8}=0\)

\(b.\left(-\dfrac{3}{4}+\dfrac{2}{5}\right):\dfrac{3}{7}+\left(\dfrac{3}{5}+-\dfrac{1}{4}\right):\dfrac{3}{7}\)

\(=\left(-\dfrac{7}{20}\right):\dfrac{3}{7}+\dfrac{7}{20}:\dfrac{3}{7}\)

\(=\left[\left(-\dfrac{7}{20}+\dfrac{7}{20}\right)\right]:\dfrac{3}{7}=0:\dfrac{3}{7}=0\)

\(c.-\dfrac{13}{18}.\dfrac{5}{8}+-\dfrac{5}{18}.\dfrac{2}{9}+-\dfrac{13}{18}.\dfrac{3}{8}+-\dfrac{5}{18}.\dfrac{7}{9}\)

\(=\left(\dfrac{5}{8}+\dfrac{3}{8}\right).-\dfrac{13}{18}+\left(\dfrac{2}{9}+\dfrac{7}{9}\right).-\dfrac{5}{18}\)

\(=1.-\dfrac{13}{18}+1.-\dfrac{5}{18}=-\dfrac{13}{18}+-\dfrac{5}{18}=-1\)

\(d.-\dfrac{11}{19}.\dfrac{4}{9}+\dfrac{-8}{19}.\dfrac{3}{7}+-\dfrac{11}{19}.\dfrac{5}{9}+-\dfrac{9}{19}.\dfrac{4}{7}\)

\(=\left(\dfrac{4}{9}+\dfrac{5}{9}\right).-\dfrac{11}{19}+-\dfrac{24}{133}+-\dfrac{36}{133}\)

\(=-\dfrac{11}{19}+-\dfrac{60}{133}=-\dfrac{137}{133}\)

a: =91/105+60/105-101/105

=50/105=10/21

c: \(\dfrac{3}{4}\cdot\dfrac{5}{2}\cdot\dfrac{7}{6}=\dfrac{3}{6}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{1}{2}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{35}{16}\)

d: =2-2/9

=18/9-2/9

=16/9

e: =24/36-9/36+8/36

=23/36

g: =5/2+1/2

=3

\(A=1+7+7^2+7^3+...+7^{2007}\)

\(7A=7+7^2+7^3+7^4+...+7^{2008}\)

\(7A-A=\left(7+7^2+7^3+7^4+...+7^{2008}\right)-\left(1+7+7^2+7^3+...+7^{2007}\right)\)

\(6A=7^{2008}-1\)

\(A=\frac{7^{2008}-1}{6}\)

Tương tự, \(B=\frac{4^{101}-1}{3},C=\frac{3^{101}-1}{2}\).

\(D=7+7^3+7^5+7^7+...+7^{99}\)

\(7^2.D=7^3+7^5+7^7+7^9+...+7^{101}\)

\(\left(7^2-1\right)D=\left(7^3+7^5+7^7+7^9+...+7^{101}\right)-\left(7+7^3+7^5+7^7+...+7^{99}\right)\)

\(48D=7^{101}-7\)

\(D=\frac{7^{101}-7}{48}\)

Tương tự, \(E=\frac{2^{9011}-2}{3}\)