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7 tháng 5 2018

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2015}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)

\(1-\frac{1}{x+1}=1-\frac{2015}{2016}\)

\(\frac{1}{x+1}=\frac{1}{2016}\)

\(x=2016-1\)

\(\Rightarrow x=2015\)

7 tháng 1 2020

Ta thấy các số hạng của vế trái đều có dạng \(\frac{1}{n\left(n+1\right)}\) với \(n\) là số tự nhiên.

Lại có: \(\frac{1}{n\left(n+1\right)}=\frac{\left(n+1\right)-n}{n\left(n+1\right)}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n+1}=\frac{1}{n}-\frac{1}{n+1}\)

Khi đó, phương trình trở thành:

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(x-1\right)x}+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{2015}{2016}\)

\(\Leftrightarrow\frac{1}{x+1}=1-\frac{2015}{2016}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2016}\)

\(\Leftrightarrow x+1=2016\)

\(\Leftrightarrow x=2015\)

Vậy \(x=2015\)

23 tháng 4 2016

b)

\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)

\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)

\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)

\(x-2=8\)

=> x = 10

23 tháng 4 2016

a) 

\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)

\(A=\frac{1}{2016}\)

4 tháng 12 2015

\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+\right)\left(x+3\right)}+...+\frac{1}{\left(x+2015\right)\left(x+2016\right)}=\frac{1}{x+2016}\)

\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+2015}-\frac{1}{x+2016}=\frac{1}{x+2016}\)

\(\frac{1}{x}-\frac{1}{x+2016}=\frac{1}{x+2016}\)

\(\frac{1}{x}-\frac{1}{x+2016}-\frac{1}{x+2016}=0\)

\(\frac{1}{x}-\frac{2x}{x+2016}=0\)

\(\frac{x+2016}{x\left(x+2016\right)}-\frac{2x}{x\left(x+2016\right)}=0\)

\(\frac{x+2016-2x}{x\left(x+2016\right)}=0\Leftrightarrow2016-x=0\Leftrightarrow x=2016\)

7 tháng 6 2019

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}\)

\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)

\(\Rightarrow2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)

\(\Rightarrow2\cdot\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2016}\div2\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{4032}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4032}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{4032}\)

\(\Rightarrow x+1=4032\Rightarrow x=4031\)

Vậy \(x=4031\)

7 tháng 6 2019

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2015}{2016}\)

=> \(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2015}{2016}\)

=> \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.x+1}\right)=\frac{2015}{2016}\)

=> \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)

=> \(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)

=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2016}:2\)

=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2032}\)

=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{2032}\)

=> \(\frac{1}{x+1}=\frac{1}{2032}\)

Vì 1 = 1

=> x + 1 = 2032

=> x = 2032 - 1

=> x = 2031

tớ ko bt lm abc , tớ lm d thôi nha , thứ lỗi 

\(\frac{5}{2x-3}-\frac{1}{x+2}=\frac{5}{x-6}-\frac{7}{2x-1}\)

\(\frac{3x+13}{2x^2+x-6}=\frac{5}{x-6}+\frac{7}{1-2x}\)

\(\frac{3x+13}{\left(x+2\right)\left(2x-3\right)}=\frac{3x+37}{\left(x-6\right)\left(2x-1\right)}\)

\(\frac{10-9x}{-4x^3+32x^2-51x+18}=0\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{10}{9}\end{cases}}\)

9 tháng 8 2016

bó tay