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9 tháng 9 2017

\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}\)\(=\frac{24}{50}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x.1}\)=\(\frac{24}{50}\)

=\(\frac{1}{2}-\frac{1}{x.1}=\frac{24}{50}\)

=\(\frac{1}{x.1}=\frac{1}{2}-\frac{24}{50}\)

=\(\frac{1}{x.1}=\frac{1}{50}\)

\(\Rightarrow\)\(x.1=50\)

\(\Rightarrow x=50\)

7 tháng 8 2016

Ta có: 1/1x2 + 1/2x3 + 1/3x4 +...+ 1/X x (X + 1) = 499/500

=> 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/X - 1/(X + 1) = 499/500

=> 1 - 1/(X + 1) = 499/500

=>      1/(X + 1) = 1 - 499/500

=>      1/(X + 1) = 1/500

=>          X + 1 = 500

=>          X       = 500 - 1

=>          X       = 499 

Đáp số: X = 499

6 tháng 4 2019

\(\left(\frac{1}{2}-\frac{3}{4}\right):\frac{1}{2}+\left(\frac{1}{2}\right)^2\)

\(=\left(\frac{2}{4}-\frac{3}{4}\right):\frac{1}{2}+\frac{1}{4}\)

\(=\frac{-1}{4}:\frac{1}{2}+\frac{1}{4}\)

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

\(=\frac{-1}{2}+\frac{1}{4}\)

\(=\frac{-2}{4}+\frac{1}{4}\)

\(=\frac{-1}{4}\)

10 tháng 5 2016

\(1-\left(\frac{73}{8}+X-\frac{173}{24}\right):\frac{50}{3}=0\)

\(\Leftrightarrow\left(\frac{73}{8}+X-\frac{173}{24}\right):\frac{50}{3}=1-0\)

\(\Leftrightarrow\left(\frac{73}{8}+X-\frac{173}{24}\right):\frac{50}{3}=1\)

\(\Leftrightarrow\left(\frac{73}{8}+X-\frac{173}{24}\right)=1.\frac{50}{3}\)

\(\Leftrightarrow\left(\frac{73}{8}+X-\frac{173}{24}\right)=\frac{50}{3}\)

\(\Leftrightarrow\frac{73}{8}+X=\frac{50}{3}+\frac{173}{24}\)

\(\Leftrightarrow\frac{73}{8}+X=\frac{191}{8}\)

\(\Leftrightarrow X=\frac{191}{8}-\frac{73}{8}\)

\(\Leftrightarrow X=\frac{59}{4}\)

Dấu \(.\)là dấu X đấy ngen pn!!!!!Nhớ K cko mik nka

10 tháng 9 2017

\(G=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

\(G=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)

\(3G=3+1+\frac{1}{3}+...+\frac{1}{3^4}\)

\(3G-G=\left(3+1+...+\frac{1}{3^4}\right)-\left(1+\frac{1}{3}+...+\frac{1}{3^5}\right)\)

\(2G=3-\frac{1}{3^5}\)

\(2G=3-\frac{1}{243}\)

\(2G=\frac{729}{243}-\frac{1}{243}\)

\(G=\frac{728}{243}:2\)

\(G=\frac{364}{243}\)

\(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{x.\left(x+1\right)}=\frac{6042}{2015}\)

\(3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{6042}{2015}\)

\(1-\frac{1}{x+1}=\frac{6042}{2015}:3\)

\(1-\frac{1}{x-1}=\frac{2014}{2015}\)

\(\frac{1}{x-1}=1-\frac{2014}{2015}\)

\(\frac{1}{x-1}=\frac{1}{2015}\)

\(\Rightarrow x-1=2015\)

\(\Rightarrow x=2016\)

31 tháng 7 2018

\(\frac{1}{3}\) + \(\frac{5}{6}\)\(\left(x-2\frac{1}{5}\right)\)\(\frac{3}{4}\)

<=> \(\frac{5}{6}\):\(\left(x-2\frac{1}{5}\right)\)\(\frac{3}{4}\)\(\frac{1}{3}\)

<=> \(\frac{5}{6}\) : \(\left(x-2\frac{1}{5}\right)\) = \(\frac{5}{12}\)

<=> \(\left(x-2\frac{1}{5}\right)\) =    \(\frac{5}{6}\) : \(\frac{5}{12}\)

,<=> \(\left(x-2\frac{1}{5}\right)\)=   2 

<=. x = 2 + \(\frac{11}{5}\)

<=> x = \(\frac{21}{5}\)

9 tháng 9 2017

Ta có:

\(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{x.\left(x+4\right)}=\frac{5}{63}\)

\(=\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{x.\left(x+4\right)}\right)=\frac{5}{63}\)

\(\Rightarrow\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+....+\frac{1}{x}-\frac{1}{x+4}\right)=\frac{5}{63}:\frac{1}{4}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{x+4}=\frac{20}{63}\Leftrightarrow\frac{1}{x+4}=\frac{1}{63}\Leftrightarrow x=63-4=59\)

12 tháng 3 2017

\(\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{n\times\left(n+1\right)}=\frac{49}{100}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{49}{100}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{n+1}=\frac{49}{100}\)

\(\Rightarrow\frac{n+1-2}{2\left(n+1\right)}=\frac{49}{100}\)

\(\Rightarrow\frac{n-1}{2n+2}=\frac{49}{100}\)

\(\Rightarrow100\left(n-1\right)=49\left(2n+2\right)\)

\(\Rightarrow100n-100=98n+98\)

\(\Rightarrow2n=198\)

=> n = 99

Vậy n =  99

12 tháng 3 2017

\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+....+\(\frac{1}{n}\)-\(\frac{1}{n+1}\)=\(\frac{49}{100}\)

\(\frac{1}{2}\)-\(\frac{1}{n+1}\)=\(\frac{49}{100}\)

         \(\frac{1}{n+1}\)=\(\frac{1}{2}\)-\(\frac{49}{100}\)

          \(\frac{1}{n+1}\)=\(\frac{1}{100}\)

=> n+1=100

        n=100-1

       n=99

7 tháng 8 2016

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{499}{500}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{499}{500}\)

\(1-\frac{1}{x+1}=\frac{499}{500}\)

\(\frac{1}{x+1}=1-\frac{499}{500}=\frac{1}{500}\)

=> x + 1 = 500

=> x = 500 - 1

=> x = 499

Vậy x = 499

7 tháng 8 2016

1/1.2 + 1/2.3 + 1/3.4 +...+ 1/x.(x+1)=499/500

1 - 1/2 + 1/2 -1/3 + 1/3 - 1/4 +...+ 1/x -1/(x+1) =499/500

1-1/(x+1)=499/500

=>x/(x+1)=499/500

=>x=499

4 tháng 5 2016

= 1/2 . 2/3 .... 2014/2015 . 2015/2016

= 1/2016

4 tháng 5 2016

1/2016