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20 tháng 8 2023

Sửa đề

\(\dfrac{2}{1^2}\cdot\dfrac{6}{2^2}\cdot\dfrac{12}{3^3}\cdot.......\cdot\dfrac{110}{10^2}\cdot x=-20\)

\(\dfrac{2}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\cdot\cdot\cdot\dfrac{11\cdot10}{10\cdot10}\cdot x=-20\)

\(\dfrac{\left(2\cdot3\cdot4\cdot....\cdot11\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot\dfrac{\left(1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot10\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot x=-20\)

\(11\cdot x=-20\\ x=-\dfrac{20}{11}\)

20 tháng 8 2023

Ko đề cho thêm \(\dfrac{20}{4²}\) mà 

10 tháng 4 2021

\(\dfrac{6}{13}.\dfrac{8}{7}+\dfrac{6}{13}.\dfrac{9}{7}-\dfrac{3}{13}.\dfrac{6}{7}=\dfrac{6}{13}.\left(\dfrac{8}{7}+\dfrac{9}{7}\right)-\dfrac{3}{13}.\dfrac{6}{7}=\dfrac{6}{13}.\dfrac{17}{7}-\dfrac{3}{13}.\dfrac{6}{7}=\dfrac{102}{91}-\dfrac{18}{91}=\dfrac{12}{13}\)

10 tháng 4 2021

Cảm ơn bạn nhiều nhéhehe

15 tháng 8 2017

a, \(\dfrac{9}{18}-\dfrac{-7}{12}+\dfrac{13}{32}\)

\(=\dfrac{1}{2}+\dfrac{7}{12}+\dfrac{13}{32}\)

\(=\dfrac{13}{12}+\dfrac{13}{32}=\dfrac{143}{96}\)

b, \(\dfrac{5}{-8}+\dfrac{14}{39}-\dfrac{6}{10}\)

\(\dfrac{-5}{8}+\dfrac{14}{39}-\dfrac{3}{5}\)

\(=\dfrac{-5}{8}-\dfrac{3}{5}+\dfrac{14}{39}\)

\(=\dfrac{-49}{40}+\dfrac{14}{39}=\dfrac{-1351}{1560}\)

16 tháng 4 2022

= 28/15 . 3/4 - ( 11/20 + 1/4 ) : 7/3

= 28/15 . 3/4 - 4/5 : 7/3

= 7/5 - 12/35

= 37/35

16 tháng 4 2022

\(\dfrac{28}{15}\) . \(\dfrac{3}{4}\) - (\(\dfrac{11}{20}\) + \(\dfrac{1}{4}\)) : \(\dfrac{7}{3}\)

\(\dfrac{7}{5}\) - (\(\dfrac{11}{20}\) + \(\dfrac{5}{20}\)) : \(\dfrac{7}{3}\)

=  \(\dfrac{7}{5}\) - \(\dfrac{16}{20}\) : \(\dfrac{7}{3}\)

=  \(\dfrac{7}{5}\) - \(\dfrac{16}{20}\) x \(\dfrac{3}{7}\)

=  \(\dfrac{7}{5}\) - \(\dfrac{12}{35}\)

=  \(\dfrac{49}{35}\) - \(\dfrac{12}{35}\)

\(\dfrac{37}{35}\)

\(\dfrac{4}{3.5}+\dfrac{8}{5.9}+\dfrac{12}{9.15}+...+\dfrac{32}{x\left(x+16\right)}=\dfrac{16}{15}\)

\(2.\left(\dfrac{2}{3.5}+\dfrac{4}{5.9}+\dfrac{6}{9.15}+..+\dfrac{16}{X.\left(X+16\right)}\right)=\dfrac{16}{15}\)

\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{15}+...+\dfrac{1}{X}-\dfrac{1}{X+16}=\dfrac{8}{15}\)

\(\dfrac{1}{X+16}=\dfrac{1}{3}-\dfrac{8}{15}\)

\(\dfrac{1}{X+16}=\dfrac{-1}{5}\)

\(X+16=-5\)

\(X=-21\)

 

DT
18 tháng 10 2023

\(\dfrac{1}{5}\left(x+\dfrac{1}{5}\right)+\dfrac{2}{5}\left(x+\dfrac{5}{3}\right)=\dfrac{98}{75}\\ =>\dfrac{1}{5}x+\dfrac{1}{25}+\dfrac{2}{5}x+\dfrac{2}{3}=\dfrac{98}{75}\\ =>\dfrac{3}{5}x=\dfrac{98}{75}-\dfrac{2}{3}-\dfrac{1}{25}=\dfrac{3}{5}\\ =>x=1\)

18 tháng 10 2023

\(\dfrac{1}{5}\left(x+\dfrac{1}{5}\right)+\dfrac{2}{5}\left(x+\dfrac{5}{3}\right)=\dfrac{98}{75}\\ \Rightarrow\dfrac{1}{5}x+\dfrac{1}{25}+\dfrac{2}{5}x+\dfrac{2}{3}=\dfrac{98}{75}\\ \Rightarrow\left(\dfrac{1}{5}x+\dfrac{2}{5}x\right)+\left(\dfrac{1}{25}+\dfrac{2}{3}\right)=\dfrac{98}{75}\\ \Rightarrow\dfrac{3}{5}x+\dfrac{53}{75}=\dfrac{98}{75}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{98}{75}-\dfrac{53}{75}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{45}{75}=\dfrac{3}{5}\\ \Rightarrow x=\dfrac{3}{5}:\dfrac{3}{5}\\ \Rightarrow x=1\)

4 tháng 4 2017

ta có

x-1/12+x-1/20+x-1/30+x-1/42+x-1/56+x-1/72=16/9

=>x-1(1/12+1/20+1/30+1/42+1/56+1/72)=16/9

=>x-1(1/3*4+1/4*5+1/5*6+1/6*7+1/7*8+1/8*9)=16/9

=>x-1(1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9)=16/9

=>x-1*(1/3-1/9)=16/9

=>(x-1)*2/9=16/9

=>x-1=9

=>x=8

kb và like cho mình nhé hihi

6 tháng 4 2017

này hà nhi đề ni dễ rữa mà ko bt

\(y+30\%y=-1,3\\ 130\%y=-1,3\\ \Rightarrow y=\dfrac{-1,3}{130\%}=-1\)

\(x:\dfrac{4}{28}=\dfrac{13}{-19}+\dfrac{8}{25}\\ 7x=-\dfrac{173}{475}\\ x=-\dfrac{\dfrac{173}{475}}{7}=-\dfrac{173}{3325}\)

20 tháng 8 2023

\(\dfrac{x}{9}=\dfrac{3}{y}+\dfrac{1}{18}\left(y\ne0\right)\)

\(\Rightarrow\dfrac{2xy}{18y}=\dfrac{54}{18y}+\dfrac{y}{18y}\)

\(\Rightarrow2xy=54+y\)

\(\Rightarrow2xy-y=54\)

\(\Rightarrow xy-\dfrac{y}{2}=27\)

\(\Rightarrow y\left(x-\dfrac{1}{2}\right)=27\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right);y\in\left\{1;3;9;27\right\}\)

\(\Rightarrow\left(x;\right)y\in\left\{\left(\dfrac{1}{2};27\right);\left(\dfrac{5}{2};9\right);\left(\dfrac{17}{2};3\right);\left(\dfrac{53}{2};1\right)\right\}\)

\(\Rightarrow\left(x;y\right)\in\varnothing\left(x;y\inℕ\right)\)