\(\frac{1}{3x4}xX+\frac{1}{4x5}xX+...+\frac{1}{49x50}xX=1\)
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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
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
NT
5
9 tháng 5 2019
\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{49.50}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+...+\left(\frac{1}{49}-\frac{1}{49}\right)-\frac{1}{50}\)
\(=\frac{1}{2}-\frac{1}{50}=\frac{12}{25}\)
~ Hok tốt ~
NH
9 tháng 5 2019
\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{49.50}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{50}\right)=2.\frac{12}{25}=\frac{24}{25}\)
I
1
KB
30 tháng 3 2022
\(lim\left(50.\dfrac{1-\left(\dfrac{4}{5}\right)^n}{1-\dfrac{4}{5}}+\dfrac{4}{5}.50.\dfrac{1-\left(\dfrac{4}{5}\right)^{n-1}}{1-\dfrac{4}{5}}\right)\) \(=50.\dfrac{1}{\dfrac{1}{5}}+\dfrac{4}{5}.50.\dfrac{1}{\dfrac{1}{5}}=450\)
10 tháng 9 2017
\(G=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(G=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
\(3G=3+1+\frac{1}{3}+...+\frac{1}{3^4}\)
\(3G-G=\left(3+1+...+\frac{1}{3^4}\right)-\left(1+\frac{1}{3}+...+\frac{1}{3^5}\right)\)
\(2G=3-\frac{1}{3^5}\)
\(2G=3-\frac{1}{243}\)
\(2G=\frac{729}{243}-\frac{1}{243}\)
\(G=\frac{728}{243}:2\)
\(G=\frac{364}{243}\)
\(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{x.\left(x+1\right)}=\frac{6042}{2015}\)
\(3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{6042}{2015}\)
\(1-\frac{1}{x+1}=\frac{6042}{2015}:3\)
\(1-\frac{1}{x-1}=\frac{2014}{2015}\)
\(\frac{1}{x-1}=1-\frac{2014}{2015}\)
\(\frac{1}{x-1}=\frac{1}{2015}\)
\(\Rightarrow x-1=2015\)
\(\Rightarrow x=2016\)
21 tháng 7 2017
\(x\cdot\frac{1}{2}\cdot\frac{1}{3}=\frac{3}{4}\\ x\cdot\frac{1}{6}=\frac{3}{4}\\ x=\frac{3}{4}:\frac{1}{6}\\ x=\frac{9}{2}\)
Vậy.....
21 tháng 7 2017
\(x\times\frac{1}{2}\times\frac{1}{3}=\frac{3}{4}\)
\(x=\frac{3}{4}\div\frac{1}{3}\div\frac{1}{2}\)
\(x=\frac{9}{2}\)
hình như là 6666666.6
Ta có:
\(\frac{1}{3.4}.x+\frac{1}{4.5}.x+...+\frac{1}{49.50}.x=1\)
\(\Rightarrow x.\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)=1\)
\(\Rightarrow x.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)=1\)
\(\Rightarrow x.\left(\frac{1}{3}-\frac{1}{50}\right)=1\Leftrightarrow x.\frac{47}{150}=1\)
\(\Rightarrow x=1:\frac{47}{150}\Leftrightarrow x=\frac{150}{47}\)