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26 tháng 7 2018

a) 24,63

26 tháng 7 2018

\(a,16,88+9,76+3,12-5,13\)

\(=\left(16,88+3,12\right)+\left(9,76-5,13\right)\)

\(=20+4,63\)

\(=24,63\)

\(b,\frac{19}{11}+\frac{8}{13}+\frac{3}{11}\)

\(=\left(\frac{19}{11}+\frac{3}{11}\right)+\frac{8}{13}\)

\(=2+\frac{8}{13}\)

\(=\frac{34}{13}\)

\(c,\frac{1}{2}\div\frac{1}{4}-\frac{1}{6}\)

\(=2-\frac{1}{6}\)

\(=\frac{11}{6}\)

25 tháng 3 2017

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

1 tháng 4 2019

1. a) \(\frac{-2}{7}+\frac{15}{23}+\frac{\left(-15\right)}{17}+\frac{4}{19}+\frac{8}{23}\)

    \(=\left(\frac{-2}{7}+\frac{-5}{7}\right)+\left(\frac{15}{23}+\frac{8}{23}\right)+\frac{4}{19}\)

    \(=\left(-1\right)+1+\frac{4}{19}\)

    \(=0+\frac{4}{19}=\frac{4}{19}\)

b) \(\frac{7}{19}\cdot\frac{8}{11}+\frac{7}{19}\cdot\frac{3}{11}+\frac{12}{19}\)

   \(=\frac{7}{19}\cdot\left(\frac{8}{11}+\frac{3}{11}\right)+\frac{12}{19}\)

   \(=\frac{7}{19}\cdot1+\frac{12}{19}\)

   \(=\frac{7}{19}+\frac{12}{19}=\frac{19}{19}=1\)

1 tháng 4 2019

2. a) \(\frac{1}{3}+\frac{\left(-2\right)}{16}-\frac{7}{14}\)

\(=\frac{5}{24}-\frac{1}{2}\)

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

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

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

\(=14+\left(-\frac{10}{91}\right)\)

\(=-14\frac{10}{91}\)

c) \(0,7\cdot2\frac{2}{3}\cdot20\cdot0,375\cdot\frac{5}{28}\)

\(=\frac{7}{10}\cdot\frac{8}{3}\cdot20\cdot\frac{3}{8}\cdot\frac{5}{28}\)

\(=\left(\frac{7}{10}\cdot\frac{5}{28}\right)\cdot\left(\frac{8}{3}\cdot\frac{3}{8}\right)\cdot20\)

\(=\frac{1}{8}\cdot1\cdot20\)

\(=\frac{20}{8}=\frac{5}{2}\)

d) \(\frac{6}{7}+\frac{5}{7}:5-\frac{8}{9}\)

\(=\frac{6}{7}+\frac{1}{7}-\frac{8}{9}\)

\(=1-\frac{8}{9}\)

\(=\frac{1}{9}\)

~Học tốt~

3 tháng 5 2019

Giúp Mik ik mai nộp oy

29 tháng 7 2019

\(a,4\frac{5}{9}:\frac{\left(-5\right)}{7}+\frac{4}{9}:\frac{-5}{7}\)

\(=\frac{41}{9}.\frac{-7}{5}+\frac{4}{9}.\frac{-7}{5}\)

\(=\frac{-7}{5}.\left(\frac{41}{9}+\frac{4}{9}\right)\)

\(=-\frac{7}{9}.5\)

\(=-7\)

a)Bn Kaito Kid làm rùi!

B)Không viết lại đề

\(=\frac{11}{7}\cdot\left(-\frac{3}{5}+\frac{4}{9}-\frac{2}{5}+\frac{5}{9}\right)=\frac{11}{7}\cdot0=0\)

c)Không viết lại đề

\(A=\left(2+4+...+100\right)\left(\frac{3}{5}\cdot\frac{10}{7}-\frac{6}{7}\right):\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(2+4+6+...+100\right)\cdot0\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)=0\)

\(=\frac{7}{6}\cdot\left(\frac{3}{26}-\frac{3}{13}+\frac{1}{10}-\frac{8}{5}\right)=\frac{7}{6}\left(\frac{-3}{26}+\frac{-17}{10}\right)=\frac{7}{6}\cdot\frac{236}{130}=\frac{413}{195}\)

D)

14 tháng 9 2016

b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=\frac{-2}{3}\)

d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}=\frac{2}{13}\)

15 tháng 9 2016

Làm tiếp:

\(=\left(1+\frac{1}{2}+.....+\frac{1}{2017}\right)-\left(1+\frac{1}{2}+....+\frac{1}{1008}\right)\)

\(=\frac{1}{1009}+\frac{1}{1010}+.........+\frac{1}{2017}\)

\(\Rightarrow\frac{\frac{1}{1009}+....+\frac{1}{2017}}{1-\frac{1}{2}+.....+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}}=1\)

Bài 2:

Đặt \(A=\frac{1}{2^2}+.......+\frac{1}{2^{800}}\)

\(4A=1+\frac{1}{2^2}+.....+\frac{1}{2^{798}}\)

\(\Rightarrow4A-A=1-\frac{1}{2^{800}}\)

\(\Rightarrow3A=1-\frac{1}{2^{800}}< 1\Rightarrow A< \frac{1}{3}\)

Vậy \(\frac{1}{2^2}+\frac{1}{2^4}+........+\frac{1}{2^{800}}< \frac{1}{3}\)

15 tháng 9 2016

Bài 1:Tính

a,   Xét biểu thức \(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).........\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)..........\left(1+\frac{n+2}{n}\right)}\) với\(n\in N\)

Ta có:\(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).......\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)......\left(1+\frac{n+2}{n}\right)}\)

\(=\frac{\frac{n+1}{1}.\frac{n+2}{2}........\frac{2n+2}{n+2}}{\frac{n+3}{1}.\frac{n+4}{2}.........\frac{2n+2}{n}}\)

\(=\frac{\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right)}{1.2.3.........\left(n+2\right)}}{\frac{\left(n+3\right)\left(n+4\right)........\left(2n+2\right)}{1.2.3.........n}}\)

\(=\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right).1.2.3.......n}{\left(n+3\right)\left(n+4\right)........\left(2n+2\right).1.2.3......\left(n+2\right)}\)

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

Áp dụng vào bài toán ta có đáp số là:1

b, \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=-\frac{2}{3}\)

c,\(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}=\frac{\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}{\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}=\frac{\frac{1}{3}}{\frac{1}{4}}=12\)

d,\(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}=\frac{2}{13}\)

e,Xét mẫu số ta có:

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)

\(=1+\frac{1}{2}-2.\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-2.\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-2.\frac{1}{2016}+\frac{1}{2017}\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{2017}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+.........+\frac{1}{2016}\right)\)