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22 tháng 8 2020

\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(A=7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{69}-\frac{1}{70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)\)

\(A=7\cdot\frac{3}{35}=\frac{21}{35}\)

22 tháng 8 2020

\(A=\frac{7}{10\cdot11}+\frac{7}{11\cdot12}+\frac{7}{12\cdot13}+...+\frac{7}{69\cdot70}\)

\(A=7\left(\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+\frac{1}{12\cdot13}+...+\frac{1}{69\cdot70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)=7\cdot\frac{3}{35}=\frac{3}{5}\)

\(B=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\)

\(B=\frac{1}{2}\left(\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+\frac{2}{29\cdot31}+...+\frac{2}{73\cdot75}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)

\(C=\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{2008\cdot2010}\)

\(C=\frac{4}{2}\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right)\)

\(C=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(C=2\left(\frac{1}{2}-\frac{1}{2010}\right)=2\cdot\frac{502}{1005}=\frac{1004}{1005}\)

3 tháng 7 2018

1) \(A=\frac{7}{10\times11}+\frac{7}{11\times12}+\frac{7}{12\times13}+...+\frac{7}{69\times70}\)

    \(A=7\times\left(\frac{1}{10\times11}+\frac{1}{11\times12}+\frac{1}{12\times13}+...+\frac{1}{69\times70}\right)\)

    \(A=7\times\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)

    \(A=7\times\left(\frac{1}{10}-\frac{1}{70}\right)\)

   \(A=7\times\frac{3}{35}\)

   \(A=\frac{3}{5}\)

2) \(B=\frac{1}{25\times27}+\frac{1}{27\times29}+\frac{1}{29\times31}+...+\frac{1}{73\times75}\)

    \(B=\frac{1}{2}\times\left(\frac{2}{25\times27}+\frac{2}{27\times29}+\frac{2}{29\times31}+...+\frac{2}{73\times75}\right)\).

    \(B=\frac{1}{2}\times\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)

    \(B=\frac{1}{2}\times\left(\frac{1}{25}-\frac{1}{75}\right)\)

    \(B=\frac{1}{2}\times\frac{2}{75}\)

    \(B=\frac{1}{75}\)

3) \(C=\frac{4}{2\times4}+\frac{4}{4\times6}+\frac{4}{6\times8}+...+\frac{4}{2008\times2010}\)

    \(C=\frac{4}{2}\times\left(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+...+\frac{2}{2008\times2010}\right)\)

    \(C=2\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

    \(C=2\times\left(\frac{1}{2}-\frac{1}{2010}\right)\)

    \(C=2\times\frac{502}{1005}\)

    \(C=\frac{1004}{1005}\)

_Chúc bạn học tốt_

29 tháng 1 2017

A = 1/25x27 + 1/27x29+1/29x31 + ...+ 1/53x55

2A = 3/25x27+ 3/27x29+...+3/53x55

2A = 1/25-1/27+1/27-1/29+...+1/53-1/55

2A = 1/25-1/55

2A = 6/275

A = 6/275:2=3/275

k mk nha, mk lm vội nên coi lại chút giùm mk xem có đúng k

4 tháng 12 2023

chịu

 

17 tháng 3 2019

\(A=\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{83.85}\)

\(\Rightarrow2A=\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{83.85}\)

\(2A=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{83}-\frac{1}{85}\)

\(2A=\frac{1}{25}-\frac{1}{85}\)

\(2A=\frac{12}{425}\)

\(A=\frac{12}{425}:2\)

\(A=\frac{6}{425}\)

\(C=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)

\(C=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)

\(C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)

\(C=\frac{1}{3}-\frac{1}{21}\)

\(C=\frac{2}{7}\)

CÂU B LÀM TƯƠNG TỰ NHA 

HOK TOT

a) Ta có: \(\left(-68+42\right)-\left(5042-6068\right)-\left(-2\right)^0\)

\(=-68+42-5042+6068-1\)

\(=6000-5000-1\)

\(=999\)

b) Ta có: \(29\cdot\left(19-37\right)-19\cdot\left(29-37\right)\)

\(=29\cdot19-29\cdot37-29\cdot19+19\cdot37\)

\(=-29\cdot37+19\cdot37\)

\(=37\cdot\left(-29+19\right)\)

\(=37\cdot\left(-10\right)=-370\)

c) Ta có: \(\left(-15\right)\cdot24+15\cdot\left(-75\right)-15\)

\(=\left(-15\right)\cdot24+\left(-15\right)\cdot75+\left(-15\right)\cdot1\)

\(=\left(-15\right)\cdot\left(24+75+1\right)\)

\(=-15\cdot100=-1500\)

d) Ta có: \(\frac{1}{5}+\frac{4}{7}-\frac{11}{35}\)

\(=\frac{5}{35}+\frac{20}{35}-\frac{11}{35}\)

\(=\frac{14}{35}=\frac{2}{5}\)

e) Ta có: \(\left(13\cdot95-73\right)-\left(13\cdot45+27\right)-\left(-1\right)^{2021}\)

\(=13\cdot95-73-13\cdot45-27-\left(-1\right)\)

\(=13\left(95-45\right)-\left(73+27\right)+1\)

\(=13\cdot50-100+1\)

\(=551\)

12 tháng 4 2020

Bài 1: Tính hợp lý

a) ( - 68 + 42 ) - ( 5042 - 6068 ) - (-2)\(^0\)

= -26 - (-1026) - 1

= - 26 + 1026 - 1

= 1000 - 1

= 999

Mấy câu còn lại c hơi làm biếng, sr :<<

 1) 9-25=(7-x)-(25+7)

(7-x)-(25+7)=9-25

(7-x)-(25+7)=-16

(7-x)-32=-16

7-x=-16+32

7-x=16

x=7-16

x=-9

2) (27-514) -(486-73)+x=7

 (27-514) -(486-73)+x=7

-487-413+x=7

-1011+x=7

x=7-(-1011)

x=7+1011

x=1018

3) 25+5+37-25+6-29-x=37

 25+5+37-25+6-29-x=37

67-25+6-29-x=37

42+6-29-x=37

48-29-x=37

19-x=37

x=19-37

x=-18

4) 14+(-12)+x =10-/-15/+ /-3/

14+(-12)+x =10-15+3

14+(-12)+x =-5+3

14+(-12)+x =-2

2+x=-2

x=-2-2

x=-4

 

15 tháng 1 2016

1) - 9

2) 1018 

3) -18 

4) -4

tick nha

Bài 1: 

Ta có: \(x-35\%\cdot x=\dfrac{1}{25}\)

\(\Leftrightarrow65\%\cdot x=\dfrac{1}{25}\)

\(\Leftrightarrow x=\dfrac{1}{25}:\dfrac{13}{20}=\dfrac{1}{25}\cdot\dfrac{20}{13}=\dfrac{4}{65}\)

Vậy: \(x=\dfrac{4}{65}\)

Bài 2: 

a) Ta có: \(17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)\)

\(=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)

\(=11+\dfrac{2}{31}-\dfrac{15}{17}\)

\(=\dfrac{5366}{527}\)