\(1.\dfrac{-7}{18}+\dfrac{-5}{12}-\dfrac{-13}{18}\text{=}\left(\dfrac{-7}{18}-\dfrac{-13}{18}\right)+\dfrac{-5}{12}\text{=}\dfrac{1}{3}+\dfrac{-5}{12}\text{=}\dfrac{-1}{12}\)

\(2.\dfrac{-13}{17}+\dfrac{-13}{21}+\dfrac{-4}{17}\text{=}\left(\dfrac{-13}{17}+\dfrac{-4}{17}\right)+\dfrac{-13}{21}\text{=}-1+\dfrac{-13}{21}\text{=}\dfrac{-34}{21}\)

\(3.\dfrac{-13}{10}-\dfrac{-4}{13}+\dfrac{-11}{10}\text{=}\dfrac{-12}{5}-\dfrac{-4}{13}\text{=}\dfrac{-136}{65}\)

\(4.\dfrac{13}{17}\times\left(\dfrac{-4}{5}+\dfrac{-3}{4}\right)\text{=}\dfrac{13}{17}\times\dfrac{-31}{20}\text{=}\dfrac{-403}{340}\)

\(5.\left(\dfrac{-5}{12}\times\dfrac{-9}{20}\right)\times\dfrac{-7}{17}\text{=}\dfrac{3}{16}\times\dfrac{-7}{17}\text{=}\dfrac{-21}{272}\)

\(6.\dfrac{11}{23}\times\left(\dfrac{5}{9}+\dfrac{17}{9}-\dfrac{13}{9}\right)\text{=}\dfrac{11}{23}\times1\text{=}\dfrac{11}{23}\)

   11 - 12 + 13  - 14  + 15 - 16 + 17 - 18 + 19  -20

= (11 + 19) - (12 + 18) + (13 + 17) - (14 + 16) - (20 - 15)

= 30 - 30 + 30 -  30 - 5

= 0 + 0 - 5

= - 5 

11-12+13-14+15-16+17-18+19-20

=(11+19)-(12+18)+(13+17)-(14+16)-(20-15)

=30-30+30-30-5

=0+0-5

=5

`a)3/5+(-4/9)`

`=3/5-4/9`

`=27/45-20/45`

`=7/45`

`b)3/5+2/5 . 15/8`

`=3/5 + 30/40`

`=3/5+3/4`

`=12/20+15/20`

`=27/20`

`c)7/2 . 8/13 + 8/13 . (-5/2)`

`=8/13 . (7/2 +(-5)/2)`

`=8/13 . 1`

`=8/13`

`d)-5/17 . (-9/23)+9/23 . (-22/17) + 11 9/23`

`=-5/17 . (-9/23) + 9/23 . (-1) . 22/17 + 11 + 9/23`

`=-5/17 . (-9/23) + (-9/23) . 22/17+11+9/23`

`= -9/23 ( -5/17 + 22/17)+11+9/23`

`= - 9/23 . 1+11+9/23`

`=-9/23+11+9/23`

`=(-9/23+9/23)+11`

`=0+11`

`=11`

a: =27/45-20/45=7/45

b: =3/5+30/40

=3/5+3/4

=12/20+15/20

=27/20

c: =8/13(7/2-5/2+1)=8/13*2=16/13

d: =9/23*5/17-9/23*22/17+11+9/23

=-9/23+11+9/23

=11

56666666

\(\frac{5.2^{13}.4^{11}-16^9}{\left(3.2^{17}\right)^2}=\frac{5.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}{3^2.\left(2^{17}\right)^2}=\frac{5.2^{13}.2^{22}-2^{36}}{3^2.2^{34}}\)

\(=\frac{5.2^{35}-2^{35}.2}{3^2.2^{34}}=\frac{2^{35}\left(5-2\right)}{3^2.2^{34}}=\frac{2^{35}.3}{3^2.2^{34}}=\frac{2^{34}.2.3}{3^2.2^{34}}=\frac{2}{3}\)

11-12+13-14+15-16+17-18+19-20

=(11-12)+(13-14)+(15-16)+(17-18)+(19-20)

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

=-1.5

=-5

ta co

   11-12+13-14+15-16+17-18+19-20\

= (11-12)+(13-14)+(15-16)+(17-18)+(19-20)

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

=(-1)x5

=-5

a) A = ( 11 + 19 ) + ( 12 + 18 ) + ( 13 + 17 ) + (14 + 16) + 15 + 20

       = 30 + 30 + 30 + 30 + 15 + 20

       = 155

b)B=11+13+15+17+...+25

  B=(11+25)+(13+23)+(15+21)+(17+19)

  B=36+36+36+36

  B=36.4

  B=144

c)C=12+14+16+18+...+26

  C=(12+26)+(14+24)+(16+22)+(18+20)

 C=38+38+38+38

 C=38.4

C=152

Bài 2 : 

a, \(x=\dfrac{3}{5}-\dfrac{7}{8}=\dfrac{24-30}{40}=-\dfrac{6}{40}=-\dfrac{3}{20}\)

b, \(2x-1=-2\Leftrightarrow x=-\dfrac{1}{2}\)

a) Số số hang là : ( 20 - 11 ) : 1 + 1 = 10 ( số ) 

Tổng là : ( 20 + 11 ) x 10 : 2 = 155

b) Số số hạng là : ( 25 - 11 ) : 2 + 1 = 13 ( số )

Tổng là : ( 25 + 11 ) x 13 : 2 = 234

c) Số số hạng là : ( 26 - 12 ) : 2 + 1 = 13 ( số )

Tổng là : ( 26 + 12 ) x 13 : 2 = 247

*** câu b và c là 2 dãy số cách nhau 2 đv 

a) Số số hạng là: (20-11)+1=10(số hạng)

= 10: 2=5 (cặp)

A = (20+11).5 = 155

b) Số số hạng là: (25-11):2+1=8(số hạng)

= 8: 2 = 4 (cặp)

B= (25+11).4=144

c) Số số hạng là: (26-12):2+1 = 8 (số hạng)

= 8 : 2 = 4 (cặp)

C = (26+12).4 = 152

a, A =(20+11) x10:2=155  (ssh = 10)

b, B =(25+ 11) x8 :2 = 144 (ssh =8)

c, C = (26 + 12 ) x8 :2 =152 (ssh = 8)

\(\text{a)}A=\frac{9}{11}+\frac{5}{7}-\frac{20}{11}+\frac{8}{13}+\frac{2}{7}\)

  \(A=\frac{9}{11}+\frac{5}{7}+\frac{-20}{11}+\frac{8}{13}+\frac{2}{7}\)

  \(A=\left(\frac{9}{11}+\frac{-20}{11}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)+\frac{8}{13}\)

  \(A=\frac{-11}{11}+\frac{7}{7}+\frac{8}{13}\)

  \(A=\left[\left(-1\right)+1\right]+\frac{8}{13}\)

\(A=0+\frac{8}{13}=\frac{8}{13}\)

\(\text{b)}B=\frac{8}{13}+\frac{9}{-17}+\frac{-34}{13}+\frac{-8}{17}\)

   \(B=\left(\frac{8}{13}+\frac{-34}{13}\right)+\left(\frac{-9}{17}+\frac{-8}{17}\right)\)

   \(B=\frac{-26}{13}+\left(-1\right)\)

   \(B=\left(-2\right)+\left(-1\right)=-3\)

\(\text{Hok tốt!}\)

\(\text{@Kaito Kid}\)