a) \(\frac{10}{17}-\frac{5}{13}+\frac{7}{17}+\frac{-8}{13}-\frac{11}{25}\)

\(=\frac{10}{17}+\frac{-5}{13}+\frac{7}{17}+\frac{-8}{13}+\frac{-11}{25}\)

\(=\left(\frac{10}{17}+\frac{7}{17}\right)+\left(\frac{-5}{13}+\frac{-8}{13}\right)+\frac{-11}{25}\)

\(=1+\left(-1\right)+\frac{-11}{25}\)

\(=0+\frac{-11}{25}\)

\(=\frac{-11}{25}\)

a) \(\frac{10}{17}-\frac{5}{13}+\frac{7}{17}+\frac{-8}{13}-\frac{11}{25}\)

\(=\frac{10}{17}+\frac{-5}{13}+\frac{7}{17}+\frac{-8}{13}+\frac{-11}{15}\)

\(=\left(\frac{10}{17}+\frac{7}{17}\right)+\left(\frac{-5}{13}+\frac{-8}{13}\right)+\frac{-11}{15}\)

\(=1+\left(-1\right)+\frac{-11}{15}\)

\(=0+\frac{-11}{15}\)

\(=\frac{-11}{15}\)

b) 1 - 2 + 3 - 4 + ...... + 2011 - 2012 ( có 2012 số )

= ( 1 - 2 ) + ( 3 - 4 ) + ....... + ( 2011 - 2012 ) ( có 1006 nhóm )

= ( - 1 ) + ( - 1 ) + ........ + ( - 1 ) ( có 1006 số )

= ( - 1 ) . 1006

= - 1006

Ta có:\(\frac{8^2}{7.9}.\frac{9^2}{8.10}.\frac{10^2}{9.11}...\frac{14^2}{13.15}=\frac{8^2.9^2.....14^2}{7.9.8.10.9.11....13.15}\)

\(=\)\(\frac{\left(8.9.10...14\right)\left(8.9.10...14\right)}{\left(7.8.9...13\right).\left(9.10.11...15\right)}\)

\(=\frac{14.8}{7.15}=\frac{2.7.8}{7.15}=\frac{2.8}{15}=\frac{16}{15}\)

\(\frac{8^2}{7.9}.\frac{9^2}{8.10}.\frac{10^2}{9.11}.\frac{11^2}{10.12}.\frac{12^2}{11.13}.\frac{13^2}{12.14}.\frac{14^2}{13.15}\)

\(\frac{8^2.9^2.10^2.11^2.12^2.13^2.14^2}{7.9.8.10.9.11.10.12.11.13.12.14.13.15}\)

\(\frac{8.9.10.11.12.13.14}{7.9.10.11.12.13.15}=\frac{8.14}{7.15}=\frac{112}{105}=\frac{16}{15}\)

Học tốt@_@

\(\frac{58}{89}=\frac{58.18}{89.18}=\frac{1044}{1602}\)

\(\frac{36}{53}=\frac{36.29}{53.29}=\frac{1044}{1537}\)

Nhận thấy: 1602 > 1537 => \(\frac{1044}{1602}< \frac{1044}{1537}\)

=> \(\frac{58}{89}< \frac{36}{53}\)

ko quy đồng mà

       \(13x+\left(-45\right)=11x-23\)

\(\Leftrightarrow11x+2x-45=11x-23\)

\(\Leftrightarrow2x-45=-23\)

\(\Leftrightarrow2x=-23+45\)

\(\Leftrightarrow2x=22\)

\(\Leftrightarrow x=22\div2\)

\(\Rightarrow x=11\)

\(Happy\)\(new\)\(year\)\(2018!\)

k đúng cho mình nhé bạn

13x+(-45)=11x-23

(13x-12x)-45=-23

2x-45=-23

2x=-23+45

2x=22

x=22:2

Vậy x=11

              CHÚC BẠN HỌC TỐT!

Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\) ( loại bỏ những phân số đối nhau ) 

\(A=1-\frac{1}{n+1}\)

\(A=\frac{n+1}{n+1}-\frac{1}{n+1}\)

\(A=\frac{n+1-1}{n+1}\)

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

Vậy \(A=\frac{n}{n+1}\)

Chúc bạn học tốt ~

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{n\cdot\left(n+1\right)}\)

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

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

chúc bạn học tốt@_@

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{n\left(n+2\right)}\)

\(=\frac{1}{2}\left(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{n}-\frac{2}{n+2}\right)\)

\(=\frac{1}{2}\left(2-\frac{2}{n+2}\right)=\frac{1}{2}\cdot\frac{2n+2}{n+2}=\frac{n+1}{n+2}< \frac{2003}{2004}\)

\(\Rightarrow\hept{\begin{cases}n+1=2002\\n+2=2003\end{cases}}\Leftrightarrow n=2001\)