13095579

dễ mà

a: \(\left(5+\frac15-\frac29\right)-\left(2-\frac{1}{23}-\frac{3}{35}+\frac56\right)-\left(8+\frac27-\frac{1}{18}\right)\)

\(=5+\frac15-\frac29-2+\frac{1}{23}+\frac{3}{35}-\frac56-8-\frac27+\frac{1}{18}\)

\(=\left(5-2-8\right)+\left(\frac15+\frac{3}{35}-\frac27\right)+\left(-\frac29-\frac56+\frac{1}{18}\right)+\frac{1}{23}\)

\(=\left(-5\right)+\left(\frac{7}{35}+\frac{3}{35}-\frac{10}{35}\right)+\left(-\frac{4}{18}-\frac{15}{18}+\frac{1}{18}\right)+\frac{1}{23}\)

\(=-5+\left(-\frac{18}{18}\right)+\frac{1}{23}=-6+\frac{1}{23}=-\frac{138}{23}+\frac{1}{23}=-\frac{137}{23}\)

c: \(-\frac57-\left(-\frac{5}{67}\right)+\frac{13}{10}+\frac12+\left(-\frac16\right)+1\frac{3}{14}-\left(-\frac25\right)\)

\(=-\frac57+\frac{5}{67}+\frac{13}{10}+\frac12-\frac16+\frac{17}{14}+\frac25\)

\(=\left(-\frac57+\frac{17}{14}+\frac12\right)+\left(\frac{13}{10}+\frac25-\frac16\right)+\frac{5}{67}\)

\(=\left(-\frac{10}{14}+\frac{17}{14}+\frac12\right)+\left(\frac{13}{10}+\frac{4}{10}-\frac16\right)+\frac{5}{67}\)

\(=\left(\frac{7}{14}+\frac12\right)+\left(\frac{17}{10}-\frac16\right)+\frac{5}{67}=1+\frac{5}{67}+\frac{51}{30}-\frac{5}{30}\)

\(=\frac{72}{67}+\frac{46}{30}=\frac{72}{67}+\frac{23}{15}=\frac{2621}{1005}\)

d: \(\frac35:\left(-\frac{1}{15}-\frac16\right)+\frac35:\left(-\frac13-1\frac{1}{15}\right)\)

\(=\frac35:\left(-\frac{2}{30}-\frac{5}{30}\right)+\frac35:\left(-\frac{5}{15}-\frac{16}{15}\right)\)

\(=\frac35:\left(-\frac{7}{30}\right)+\frac35:\left(-\frac{21}{15}\right)\)

\(=\frac35\cdot\frac{-30}{7}+\frac35\cdot\frac{-5}{7}=\frac35\cdot\left(-\frac{30}{7}-\frac57\right)\)

\(=\frac35\cdot\left(-\frac{35}{7}\right)=\frac35\cdot\left(-5\right)=-3\)

Lời giải chi tiết:

Dễ nhỉ !

Ai đồng tình thì cho mình nha !

a: =>x-2/5=3/4:1/3=3/4*3=9/4

=>x=9/4+2/5=45/20+8/20=53/20

b: =>x-2/3=7/3:4/5=7/3*5/4=35/12

=>x=35/12+2/3=43/12

c: 1/3(x-2/5)=4/5

=>x-2/5=4/5*3=12/5

=>x=12/5+2/5=14/5

d: =>2/3x-1/3-1/4x+1/10=7/3

=>5/12x-7/30=7/3

=>5/12x=7/3+7/30=77/30

=>x=77/30:5/12=154/25

e: \(\Leftrightarrow x\cdot\dfrac{3}{7}-\dfrac{2}{7}+\dfrac{1}{2}-\dfrac{5}{4}x+\dfrac{5}{2}=0\)

=>\(x\cdot\dfrac{-23}{28}=\dfrac{2}{7}-3=\dfrac{-19}{7}\)

=>x=19/7:23/28=76/23

f: =>1/2x-3/2+1/3x-4/3+1/4x-5/4=1/5

=>13/12x=1/5+3/2+4/3+5/4=257/60

=>x=257/65

i: =>x^2-2/5x-x^2-2x+11/4=4/3

=>-12/5x=4/3-11/4=-17/12

=>x=17/12:12/5=85/144

Bài 1:

\(\frac14+\frac23=\frac{3}{12}+\frac{8}{12}=\frac{3+8}{12}=\frac{11}{12}\)

\(\frac27+\frac23=\frac{6}{21}+\frac{14}{21}=\frac{6+14}{21}=\frac{20}{21}\)

\(\frac25+\frac13=\frac{6}{15}+\frac{5}{15}=\frac{6+5}{15}=\frac{11}{15}\)

\(\frac23+\frac12=\frac46+\frac36=\frac{4+3}{6}=\frac76\)

\(\frac13+\frac35=\frac{5}{15}+\frac{9}{15}=\frac{5+9}{15}=\frac{14}{15}\)
\(\frac45+\frac13=\frac{12}{15}+\frac{5}{15}=\frac{12+5}{15}=\frac{17}{15}\)

\(\frac18+\frac34=\frac18+\frac68=\frac{1+6}{8}=\frac78\)

\(\frac{1}{36}+\frac{5}{12}=\frac{1}{36}+\frac{15}{36}=\frac{1+15}{36}=\frac{16}{36}=\frac49\)

\(\frac13+\frac16+\frac{1}{18}=\frac{6}{18}+\frac{3}{18}+\frac{1}{18}=\frac{6+3+1}{18}=\frac{10}{18}=\frac59\)

Bài 2:

\(\frac{15}{16}-\frac{3}{16}=\frac{15-3}{16}=\frac{12}{16}=\frac34\)

\(\frac{17}{18}-\frac56=\frac{17}{18}-\frac{15}{18}=\frac{17-15}{18}=\frac{2}{18}=\frac19\)

\(\frac34-\frac49=\frac{27}{36}-\frac{16}{36}=\frac{27-16}{36}=\frac{11}{36}\)

\(\frac12-\frac25=\frac{5}{10}-\frac{4}{10}=\frac{5-4}{10}=\frac{1}{10}\)

\(\frac56-\frac{3}{10}=\frac{25}{30}-\frac{9}{30}=\frac{25-9}{30}=\frac{16}{30}=\frac{8}{15}\)

\(3-\frac13=\frac93-\frac13=\frac{9-1}{3}=\frac83\)

\(\frac45-\frac{1}{10}=\frac{8}{10}-\frac{1}{10}=\frac{7}{10}\)

\(\frac52-1=\frac{5-2}{2}=\frac32\)

\(\frac58-\frac25=\frac{25}{40}-\frac{16}{40}=\frac{25-16}{40}=\frac{9}{40}\)

4: Ta có: \(\dfrac{1}{3+\sqrt{5}}-\dfrac{1}{3-\sqrt{5}}\)

\(=\dfrac{3-\sqrt{5}-3-\sqrt{5}}{4}\)

\(=\dfrac{-\sqrt{5}}{2}\)

   \(\dfrac{2}{5}+\dfrac{2}{3}+\dfrac{2}{4}\)

= \(\dfrac{24}{60}\) + \(\dfrac{40}{60}\) + \(\dfrac{30}{60}\)

= \(\dfrac{64}{60}\) + \(\dfrac{30}{60}\)

= \(\dfrac{47}{30}\)

\(\dfrac{2}{6}+\dfrac{3}{12}\)

= \(\dfrac{4}{12}\) + \(\dfrac{3}{12}\)

= \(\dfrac{7}{12}\)

\(\dfrac{5}{6}\) + \(\dfrac{1}{3}\)

= \(\dfrac{5}{6}\) + \(\dfrac{2}{6}\)

= \(\dfrac{7}{6}\)

   \(\dfrac{1}{3}\) + \(\dfrac{5}{12}\) + \(\dfrac{5}{6}\)

= \(\dfrac{4}{12}\) + \(\dfrac{5}{12}\) + \(\dfrac{10}{12}\)

= \(\dfrac{9}{12}\) + \(\dfrac{10}{12}\)

= \(\dfrac{19}{12}\)

    \(\dfrac{5}{8}\) + \(\dfrac{4}{7}\)

= \(\dfrac{35}{56}\) + \(\dfrac{32}{56}\)

=  \(\dfrac{67}{56}\) 

\(\dfrac{7}{3}\) + \(\dfrac{8}{7}\)

= \(\dfrac{49}{21}\) + \(\dfrac{24}{21}\)

= \(\dfrac{73}{21}\)

   \(\dfrac{1}{5}+\dfrac{5}{35}\)

= \(\dfrac{7}{35}\) + \(\dfrac{5}{35}\)

= \(\dfrac{12}{35}\) 

Kho..................wa.....................troi.....................thi......................lanh.................ret.......................ai........................tich..........................ung.....................ho........................minh.....................cho....................do....................lanh

\(7832\)

a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

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

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

= \(1-\frac{1}{10}\)

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

b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)

=\(1-\frac{1}{11}\)

= \(\frac{10}{11}\)

c) đặt A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)

     \(\frac{1}{3}A\)=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

     \(\frac{2}{3}A\)=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

      \(\frac{2}{3}A\)=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

     \(\frac{2}{3}A\)=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)

     \(\frac{2}{3}A\)=\(\frac{10}{11}\)

         A= \(\frac{10}{11}:\frac{2}{3}\)

          A= \(\frac{10}{11}.\frac{3}{2}\)=\(\frac{15}{11}\)

d) giả tương tự câu c kết quả \(\frac{25}{11}\)

tổng đặc biệt đó bạn

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)

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

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

những câu sau cũng áp dụng như vậy nhé

hey , đề bài sai ròi .

A = \(\dfrac{1}{1+3}\) + \(\dfrac{1}{1+3+5}\) + \(\dfrac{1}{1+3+5+7}\) + ... + \(\dfrac{1}{1+3+5+7+...+2021}\)

\(\Leftrightarrow\) A = \(\dfrac{1}{\dfrac{\left(1+3\right).2}{2}}\) + \(\dfrac{1}{\dfrac{\left(1+5\right).3}{2}}\) + \(\dfrac{1}{\dfrac{\left(1+7\right).4}{2}}\) + ... + \(\dfrac{1}{\dfrac{\left(1+2021\right).1011}{2}}\)

= \(\dfrac{2}{2.4}\) + \(\dfrac{2}{3.6}\) + \(\dfrac{2}{4.8}\) + ... + \(\dfrac{2}{1011.2021}\)

= \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + \(\dfrac{1}{4.4}\) + ... + \(\dfrac{1}{2021.2021}\)

A < \(\dfrac{1}{4}\) + ( \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ... + \(\dfrac{1}{2020.2021}\) )

< \(\dfrac{1}{4}\) + ( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{2020}\) - \(\dfrac{1}{2021}\) )

< \(\dfrac{1}{4}\) + ( \(\dfrac{1}{2}\) - \(\dfrac{1}{2021}\) ) < \(\dfrac{1}{4}\) + \(\dfrac{1}{2}\) = \(\dfrac{3}{4}\)

Kiểu như vậy hả ?