1: =5/8*17/13*7/17*8/3*6

=5/3*7/13*6

=5*7*2/13=70/13

2: =-7/9(24/19-1)

=-7/9*5/19

=-35/171

3: =-3/8(7/6-9/6)

=-3/8*(-2/6)

=3/8*1/3=1/8

`a// [-3]/5+7/21+[-4]/5+7/5=([-3]/5+[-4]/5+7/5)+7/21=0/5+7/21=7/21`

`b// [-3]/17+(2/3+3/17)=[-3]/17+2/3+3/17=([-3]/17+3/17)+2/3=0/17+2/3=2/3`

`c// [-5]/21+([-16]/21+1)=[-5]/21+[-16]/21+1=([-5]/21+[-16]/21)+1=[-21]/21+1=-1+1=0`

a) \(\dfrac{-3}{5}+\dfrac{7}{21}+\dfrac{-4}{5}+\dfrac{7}{5}\)

\(=\left(-\dfrac{3}{5}+\dfrac{-4}{5}+\dfrac{7}{5}\right)+\dfrac{7}{21}\)

\(=0+\dfrac{7}{21}\)

\(=\dfrac{7}{21}\)

b) `-3/17 + ( 2/3 + 3/17)`

` = -3/17 +  2/3 + 3/17`

` = 2/3 + ( -3/17 +3/17)`

` = 2/3 + 0`

` = 2/3`

c) 

` -5/21 + ( -16/21 +1)`

\(=\dfrac{-5}{21}+\dfrac{-16}{21}+1\)

\(=\left(\dfrac{-5}{21}+\dfrac{-16}{21}\right)+1\)

\(=-1+1=0\)

Giúp tớ với tớ cần gấp

a: \(=\dfrac{-5}{7}-\dfrac{2}{7}+\dfrac{3}{4}+\dfrac{1}{4}-\dfrac{1}{5}=-\dfrac{1}{5}\)

b: \(=\dfrac{-3}{31}-\dfrac{28}{31}+\dfrac{-6}{17}-\dfrac{11}{17}+\dfrac{1}{29}-\dfrac{1}{5}=\dfrac{-24}{145}\)

Đề khó nhìn quá bạn ơi!

\(a,1-2+3-4+5-6+......+199-200\)

\(=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+.....+\left(199-200\right)\)( 100 cặp )

\(=-1+\left(-1\right)+\left(-1\right)+........+\left(-1\right)\)( 100 số hạng )

\(=-1.100\)

\(=-100\)

\(a.1-2+3-4+5-6+...+199-200\)

\(=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(199-200\right)\)  (có tất cả \(200:2=100\)cặp)

\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)

\(=\left(-1\right).200=-200\)

\(b.1+2-3-4+5+6-7-8+...+97+98-99-100\)

\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(97+98-99-100\right)\)  (có \(100:4=25\)cặp)

\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)

\(=\left(-4\right).25=-100\)

\(c.1+\left(-6\right)+11+\left(-16\right)+...+21+\left(-26\right)\)

\(=\left[1+\left(-6\right)\right]+\left[11+\left(-16\right)\right]+...+\left[21+\left(-26\right)\right]\) (có tất cả \(26:2=13\)cặp)

\(=\left(-5\right)+\left(-5\right)+...+\left(-5\right)\)

\(=-5.13=-65\)

Câu a tớ làm sai, bạn tham khảo của bạn kia nhé!

1+(-2)+3+(-4)+5+(-6)+...+21+(-22)=[1+(-2)]+[3+(-4)]+...+[21+(-22)]

                                                   -----------11 cặp----------------------

=(-1)+(-1)+...+(-1)=(-1).11=-11

=(1 - 2 - 3 + 4) + (5 - 6 - 7 + 8) + ......+ ( 21 - 22 - 23 + 24)

= 0+0+0+......+0

=0 nha

1-2-3+4+5-6-7+8+.....+21-22-23+14

=(1-2-3+4)+(5-6-7+8)+....+(21-22-23+24)

=0+0+...+0

=0

1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 +...+ 21 - 22 - 23 + 24

= (1-2-3+4) + (5-6-7+8) + .... + (21-22-23+24)

= 0 + 0 + ... + 0

= 0

Hok tốt !

Lời giải:
Áp dụng định lý Fermat nhỏ:

Với $a$ là số tự nhiên sao cho $(a,11)=1$ thì:

$a^{10}\equiv 1\pmod {11}\Rightarrow a^{3330}\equiv 1\pmod {11}$

$\Rightarrow a^{3331}\equiv a\pmod {11}$

Còn với mọi $a\vdots 11$ thì $a^{3331}\equiv a\pmod {11}$ (hiển nhiên)

Do đó:

$1^{3331}+2^{3331}+...+2020^{3331}\equiv 1+2+3+...+2020\equiv 1010.2021\equiv 9.8\equiv 6\pmod {11}$

$\Rightarrow 1^{3331}+2^{3331}+...+2020^{3331}-6\equiv 0\pmod {11}$

Ta có đpcm.