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1: 

Ta có: \(D=\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+\dfrac{3}{9\cdot11}+...+\dfrac{3}{53\cdot55}\)

\(=\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+...+\dfrac{2}{53\cdot55}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{55}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{11}{55}-\dfrac{1}{55}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{2}{11}=\dfrac{3}{11}\)

2) Để A là số nguyên dương thì 

\(\left\{{}\begin{matrix}x+2⋮x-5\\x-5>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5+7⋮x-5\\x>5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7⋮x-5\\x>5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5\inƯ\left(7\right)\\x>5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-5\in\left\{1;-1;7;-7\right\}\\x>5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{6;4;12;-2\right\}\\x>5\end{matrix}\right.\)

\(\Leftrightarrow x\in\left\{6;12\right\}\)

26 tháng 7 2016

\(\text{Ta có:}\) \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}\)

\(\Leftrightarrow2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}.2\)

\(\Leftrightarrow\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).x=\frac{4}{3}\)

\(\Leftrightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right).x=\frac{4}{3}\)

\(\Leftrightarrow\left(1-\frac{1}{11}\right)x=\frac{4}{3}\)

\(\Leftrightarrow\frac{10}{11}x=\frac{4}{3}\)

\(\Leftrightarrow x=\frac{4}{3}:\frac{10}{11}=\frac{22}{15}\)

1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)

\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)

=0

2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)

\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)

\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)

3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)

\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)

\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)

\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)

\(=\dfrac{52546}{15}\)

4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)

\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)

\(=\dfrac{4}{7}\)

2 tháng 7 2021

cảm ơn nhé

vui

10 tháng 6 2020

\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{99}\right)\)

\(=\left(\frac{2}{2}+\frac{1}{2}\right)\left(\frac{3}{3}+\frac{1}{3}\right)\left(\frac{4}{4}+\frac{1}{4}\right).....\left(\frac{99}{99}+\frac{1}{99}\right)\)

\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)

\(=\frac{3.4.5....100}{2.3.4....99}=\frac{100}{2}=50\)

25 tháng 8 2016

= 1/10

26 tháng 8 2016

bài 1 x bằng 0

bài 2 bằng 1/10

27 tháng 8 2015

Bạn gõ thiều phân số \(\frac{1}{3.5}\)à?

27 tháng 8 2015

Mik giải phía dưới rồi đó. Câu lúc nãy bạn đăng ý

27 tháng 8 2015

\(\left[\frac{12}{11}-\left(\frac{1}{2}+\frac{1}{44}\right)\right].\left(x-0,2\right)=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

\(\frac{25}{44}.\left(x-0,2\right)=\frac{1}{2}.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{9.11}\right)\)

\(x-0,2=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right):\frac{25}{44}\)

\(x-\frac{1}{5}=\frac{22}{25}.\left(1-\frac{1}{11}\right)=\frac{22}{25}.\frac{10}{11}=\frac{4}{5}\)

\(x=\frac{4}{5}+\frac{1}{5}\)

\(x=1\)