\(\frac{-5}{7}.\frac{3}{9}+\frac{4}{9}:\frac{-...">
K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

29 tháng 6 2020

c) =-5/7x3/9+4/9x3/-7

=-5/7x3/9+3/9x4/-7

=3/9x(-5/7+4/-7)

=3/9x-9/7

=-3/7

d) =58/7 - (31/9+30/7)

=58/7-31/9-30/7

=(58/7-30/7)-(31/9)

=4-31/9

=5/9

13 tháng 6 2020

(1-1/3)x(1-1/5)x(1-1/7)x(1-1/9)x(1-1/2)x(1-1/4)x(1-1/6)x(1-1/8)x(1-1/10)

=2/3x4/5x6/7x8/9x1/2x3/4x5/6x7/8x9/10

=2x4x6x8x1x3x5x7x9 /3x5x7x9x2x4x6x8x10

=1/10

13 tháng 6 2020

\(b,\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)\)

\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)

\(=\frac{100}{2}\)

\(=50\)

19 tháng 6 2016

a.. A = 310 × 11 + 310 × 5 / 39 × 24

A = 310 × (11 + 5) / 39 × 24

A = 310 × 16 / 39 × 24

A = 310 × 24 / 39 × 24

A = 3

C = 11 × 322 × 37 - 915 / (2 × 314)2

C = 11 × 329 - (32)15 / 22 × 328

C = 11 × 329 - 330 / 22 × 328

C = 329 × (11 - 3) / 22 × 328

C = 329 × 8 / 22 × 328

C = 329 × 23 / 22 × 328

C = 3 × 2

C = 6

-5/9+8/15+-2/11+-4/9/7/17

(-5/9+-4/9)+(8/15+7/15+-2/11

-1+1+-2/11

-2/11

(17/5+11/4)-22/5

123/20-22/5

123/20-88/20

35/20

1 tháng 8 2020

thanks friend!vui

a) Ta có: \(15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\)

\(=15+\frac{3}{13}-3-\frac{4}{7}-8-\frac{3}{13}\)

\(=4-\frac{4}{7}=\frac{24}{7}\)

b) Ta có: \(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)

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

\(=8+\frac{7}{11}=\frac{95}{11}\)

c) Ta có: \(\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+5\frac{7}{9}\)

\(=\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+\frac{-7}{9}\cdot\frac{-52}{7}\)

\(=\frac{-7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}-\frac{52}{7}\right)\)

\(=\frac{-7}{9}\cdot\frac{45}{-7}=5\)

d) Ta có: \(50\%\cdot1\frac{1}{3}\cdot10\cdot\frac{7}{35}\cdot0.75\)

\(=\frac{1}{2}\cdot\frac{4}{3}\cdot10\cdot\frac{7}{35}\cdot\frac{3}{4}\)

\(=5\cdot\frac{7}{35}=1\)

e) Ta có: \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)

\(=1-\frac{1}{43}=\frac{43}{43}-\frac{1}{43}\)

\(=\frac{42}{43}\)