\(A=21\frac{4}{11}-\left(1\frac{3}{5}+7\frac{4}{11}\right)\)

\(A=\frac{235}{11}-\left(\frac{8}{5}+\frac{81}{11}\right)\)

\(A=\left(\frac{235}{11}-\frac{81}{11}\right)+\frac{8}{5}\)

\(A=\frac{154}{11}+\frac{8}{5}\)

\(\Rightarrow A=\frac{78}{5}\)

\(B=\left(7\frac{8}{9}+2\frac{3}{13}\right)-\left(4\frac{8}{9}-7\frac{10}{13}\right)\)

\(B=\left(\frac{71}{9}+\frac{29}{13}\right)-\left(\frac{44}{9}-\frac{101}{13}\right)\)

\(B=\left(\frac{71}{9}-\frac{44}{9}\right)+\left(\frac{29}{13}-\frac{101}{13}\right)\)

\(B=\frac{27}{9}+\frac{-72}{13}\)

\(B=3+\frac{-72}{13}\)

\(\Rightarrow B=\frac{-33}{13}\)

P/s: Hoq chắc :v

giúp em với : Akai Haruma

A =

B =

C =

C =

C = (4.-12)/(13.17) +

C =

C =

C = 96/221

D = -1/12 - ( 2

D = -1/12 - ( 4

D = -1/12 - (-1

D = -1/12 + 1

-3/36 + 2

D = -1/36

):5/11

E = ):5/11

E = 0:5/11

E = 0

a, \(\frac{1}{4}+\frac{5}{12}-\frac{1}{13}-\frac{7}{8}\)

\(=\left(\frac{1}{4}+\frac{5}{12}\right)-\left(\frac{1}{13}+\frac{7}{8}\right)\)

\(=\frac{2}{3}-\frac{99}{104}\)

\(=-\frac{89}{312}\)

b, \(11\frac{3}{13}-2\frac{4}{7}+5\frac{3}{13}\)

\(=\left(11\frac{3}{13}+5\frac{3}{13}\right)-2\frac{4}{7}\)

\(=\frac{214}{13}-\frac{18}{7}\)

\(=\frac{1264}{91}\)

c, \(\left(6\frac{4}{9}+3\frac{7}{11}\right)-4\frac{4}{9}\)

\(=6\frac{4}{9}+3\frac{7}{11}-4\frac{4}{9}\)

\(=\left(6\frac{4}{9}-4\frac{4}{9}\right)+3\frac{7}{11}\)

\(=2+3\frac{7}{11}\)

\(=5\frac{7}{11}\)

\(=\frac{62}{11}\)

d, \(\left(6,17+3\frac{5}{9}-2\frac{36}{97}\right)\left(\frac{1}{3}-0,25-\frac{1}{12}\right)\)

\(=\left(6,17+3\frac{5}{9}-2\frac{36}{97}\right)\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)\)

\(=\left(6,17+3\frac{5}{9}-2\frac{36}{97}\right)\cdot0\)

\(=0\)

e, \(-1,5\cdot\left(1+\frac{2}{3}\right)\)

\(=-\frac{3}{2}\cdot\frac{5}{3}\)

\(=-\frac{5}{2}\)

f, Đặt \(A=1^2+2^2+3^2+...+100^2\)

\(=1+2\left(3-1\right)+3\left(4-1\right)+...+100\left(101-1\right)\)

\(=1+2\cdot3-2+3\cdot4-3+...+100\cdot101-100\)

\(=\left(2\cdot3+3\cdot4+...+100\cdot101\right)-\left(1+2+3+...+100\right)\)

Đặt B = 2 . 3 + 3 . 4 + ... + 100 . 101 

3B = 2 . 3 ( 4 - 1 ) + 3 . 4 ( 5 - 2 ) + ... + 100 . 101 . ( 102 - 99 )

3B = 2 . 3 . 4 - 1 . 2 . 3 + 3 . 4 . 5 - 2 . 3 . 4 + ... + 100 . 101 . 102 - 99 . 100 . 101 

3B = 100 . 101 . 102

B = \(\frac{100\cdot101\cdot102}{3}\)

B = 343400

Thay B vào A. Ta được :

\(A=343400-\left(1+2+3+...+100\right)\)

Thay C = 1 + 2 + 3 + ... + 100

Dãy số 1; 2; 3; ...; 100 có số số hạng là:

( 100 - 1 ) : 1 + 1 = 100 ( số hạng )

Tổng của dãy số đó là :

( 100 + 1 ) . 100 : 2 = 5050

=> C = 5050

Thay C vào A. Ta được :

\(A=343400-5050\)

\(A=338350\)

Vậy A = 338350

a) = -3/7 . 5/11 + -3/7 . 6/11 + 9/7

   = -3/7. ( 5/11 + 6/11 ) + 9/7

  = -3/7. 1 + 9/7

  = -3/7 + 9/7

  = 6/7

b) = 4/13 + 9/13 + -11/5 + 6/5 - 3/4

    = 13/13 + -5/5 - 3/4

    = 1 + (-1) - 3/4

    = 0 - 3/4

    = -3/4

c) = -19/17. 4/7 + 19/17. -3/7 + 19/17

    = 19/17. -4/7 + 19/17. -3/7 + 19/17.1

    = 19/17.( -4/7 + -3/7 + 19/17

    = 19/17. -7/7 + 19/17

    = 19/17. (-1) + 19/17

    = -19/17 + 19/17

    = 0

tk mk nha,thanks

Tính nhanh :

a) 

\(\left(2+\frac{-7}{13}\right)+\frac{-6}{13}\)

= \(2+\left(\frac{-7}{13}+\frac{-6}{13}\right)\)

= \(2+\frac{-13}{13}=2+\left(-1\right)\)

= 1

tk mk nha

chắc chắn đúng luôn ấy

mk đang âm rất nhiều này ! huhuhu giúp nha  

a)=1

b)=\(-\frac{1}{3}\)

c)=\(\frac{9}{29}\)