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13 tháng 10 2015

\(=87+\frac{23}{59}+74+\frac{17}{71}+12+\frac{36}{59}+25+\frac{54}{71}\)

=\(\left(87+74+12+25\right)+\left(\frac{23}{59}+\frac{36}{59}\right)+\left(\frac{17}{71}+\frac{54}{71}\right)\)

=\(198+1+1=200\)

28 tháng 10 2020

Mình chưa hiểu đề lắm nhưng chắc ý bạn là như này:
   \(87\frac{23}{59}+74\frac{17}{71}+12\frac{36}{59}+25\frac{5}{71}\)
\(\left(87\frac{23}{59}+12\frac{36}{59}\right)+\left(25\frac{5}{71}+74\frac{17}{71}\right)\)
\(\left(87+12\right)+\left(\frac{23}{59}+\frac{36}{59}\right)+\left(25+74\right)+\left(\frac{5}{71}+\frac{17}{71}\right)\)
\(99+\frac{59}{59}+99+\frac{22}{71}\)
\(198+1+\frac{22}{71}\)
\(199\frac{22}{71}\)
Chúc bạn học tốt ^^!

12 tháng 10 2016

cai bai nay minh thay

quen quen nhung lai ko nghi ra

chu!

chuc bn hoc gioi1

27 tháng 2 2017

Ta có : \(\frac{1}{1000}+\frac{13}{1000}+\frac{25}{1000}+.....+\frac{85}{1000}+\frac{97}{1000}\)

\(=\frac{1+13+25+......+85+97}{1000}\)

\(=\frac{441}{1000}\)

16 tháng 9 2023

\(\dfrac{1}{12}\times\dfrac{4}{5}=\dfrac{4}{60}=\dfrac{1}{15}\\ \dfrac{9}{5}:\dfrac{4}{7}=\dfrac{9}{5}\times\dfrac{7}{4}=\dfrac{63}{20}\\ 4\times\dfrac{3}{7}=\dfrac{4\times3}{7}=\dfrac{12}{7}\\ \dfrac{1}{2}:5=\dfrac{1}{2}\times\dfrac{1}{5}=\dfrac{1}{10}\)

20 tháng 7 2017

\(=1-\frac{1}{1\cdot2}+1-\frac{1}{2\cdot3}+1-\frac{1}{3\cdot4}+...+1-\frac{1}{9\cdot10}\)

\(=\left[1+1+1+...+1\right]-\left[\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\right]\)

\(=9-\left[\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right]=9-\left[1-\frac{1}{10}\right]\)

\(=9-\frac{9}{10}=\frac{81}{10}\)

15 tháng 10 2018

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)

\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)

\(=9-\left(1-\frac{1}{10}\right)\)

\(=9-\frac{9}{10}=\frac{81}{10}\)

18 tháng 6 2018

Ta có \(71+65\times4=\frac{x+140}{x}+26\)

\(\Rightarrow71+260=\frac{x+140}{x}+260\)

\(\Rightarrow\frac{x+140}{x}=71\)( triệt tiêu hai vế đi 260 )

\(\Rightarrow x+140=71x\)

\(\Rightarrow70x=140\)

\(\Rightarrow x=2\)

Vậy x = 2

30 tháng 5 2019

\(71+65.4=\frac{x+140}{x}\)\(+260\)

\(71+260=\frac{x+140}{x}+260\)

=>\(\left(x+140\right):x=71\)

=>\(x+140=71x\)

=>\(71x-x=140\)

=>\(70x=140\)

=>\(x=140:70\)

=>\(x=2\)

Vậy x = 2

CHÚC BẠN HỌC TỐT!

28 tháng 4 2018

\(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)

\(=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)

\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{!}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(=8-\frac{2}{5}=\frac{38}{5}\)

13 tháng 9 2020

 1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90
=1-1/2+1-1/6+1-1/12+1-1/20+1-1/30+1-1/42+1-1/56+1-1/72+1-1/90
=9 – (1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)
=9 – [1/(1x2)+1/(2x3)+1/(3x4)+1/(4x5)+1/(5x6)+1/(6x7)+1/(7x8)+1/(8x9)+1/(9x10)]
=9 – ( 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)
=9 – (1 – 1/10) = 9 – 9/10
= 81/10

30 tháng 6 2018

A = 54/15.18   +   54/18.21  +  .........+   54/87.90

A = 54/3 . ( 1/15.18 + 1/18.21 + ........+ 1/87.90)

A = 54/3 . ( 1/15 - 1/18 + 1/18 -1/21 + ......+ 1/87 - 1/90)

A =54/3 . ( 1/15 -1/90)

A = 54/3 . 1/18 = 1

vậy A = 1

3 tháng 9 2015

\(\frac{19}{28};\frac{80}{79};\frac{112}{111};\frac{2013}{2012}\)

\(\frac{99}{100};\frac{61}{62};\frac{43}{45};\frac{15}{17}\)