`2/(3.5)+2/(5.7)+....+2/(2015.2017)`

`=1/3-1/5+1/5-1/7+....+1/2016-1/2017`

`=1/3-1/2017=2014/6051`

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2015.2017}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)

\(=\dfrac{1}{3}-\dfrac{1}{2017}\)

\(=\dfrac{2017}{6051}-\dfrac{3}{6051}=\dfrac{2014}{6051}\)

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2015.2017}\) 

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\) 

\(=\dfrac{1}{3}-\dfrac{1}{2017}\) 

\(=\dfrac{2014}{6051}\)

\(M=1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9999}\\ =\dfrac{3}{3}+\dfrac{3}{15}+\dfrac{3}{35}+...+\dfrac{3}{9999}\\ =\dfrac{3}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\\ =\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =\dfrac{3}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{3}{2}\cdot\dfrac{100}{101}=\dfrac{150}{101}\)

\(A=x+152=\dfrac{1}{3}+152=\dfrac{457}{3}\)

A bn lướt xuống dưới mà xem cách làm 

nhưng của bn là cho 3 ra ngoài nha

Giải:

A=3/1.3+3/3.5+3/5.7+...+3/49.51

A=3/2.(2/1.3+2/3.5+2/5.7+...+2/49.51)

A=3/2.(1/1-1/3+1/3-1/5+1/5-1/7+...+1/49-1/51)

A=3/2.(1/1-1/51)

A=3/2.50/51

A=25/17

B=1/3+1/3²+1/3³+...+1/3⁸

3B=1+1/3+1/3²+...+1/3⁷

3B-B=(1+1/3+1/3²+...+1/3⁷)-(1/3+1/3²+1/3³+...+1/3⁸)

2B=1-1/3⁸

   B=1-1/3⁸ /2

Chúc bạn học tốt!

\(=3.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{47.49}\right)\)

\(=3.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}\right)\)

\(=3.\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)

\(=3.\dfrac{46}{147}\)

\(=\dfrac{46}{49}\)

\(\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{47.49}\)

=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}\right)\)

=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)

=\(\dfrac{3}{2}.\dfrac{46}{147}\)

=\(\dfrac{23}{49}\)

=3/2.(1/3-1/5+1/5-1/7+...+1/47-1/49)

=3/2.(1/3-1/49)

=3/2.46/147=23/49

=1/4+7/8=9/8

\(\dfrac{1}{4}+\dfrac{3}{4}:\dfrac{-6}{-7}=\dfrac{1}{4}+\dfrac{3}{4}:\dfrac{6}{7}=\dfrac{1}{4}+\dfrac{3}{4}\times\dfrac{7}{6}=\dfrac{1}{4}+\dfrac{7}{8}=\dfrac{9}{8}\)

ko rảnh giải chi tiết nên:

\(\dfrac{9}{8}\)

\(\dfrac{x+3}{15}=\dfrac{1}{3}\left(1\right)\\ \Leftrightarrow x+3=\dfrac{1}{3}\cdot15\\ \Leftrightarrow x+3=5\\ \Rightarrow x=5-3\\ \Rightarrow x=2\)

Vậy tập nghiệm phương trình (1) là \(\left\{2\right\}\)

(X+3)3=15x1

3X+9=15

3X=15-9

3X=6

X=6:3

X=2

\(\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{x}{15}+\dfrac{3}{15}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{1}{3}-\dfrac{3}{15}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{2}{15}\)
\(\Rightarrow x=2\)

\(M=\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}=\frac{3\left(\frac{1}{5}+\frac{1}{7}-\frac{3}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{3}{4}\) \(\frac{3}{4}\)                                                                                                          \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=2-\frac{2}{101}=\frac{200}{101}\)

\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(B=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

\(B=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(B=2.\left(\frac{1}{1}-\frac{1}{101}\right)\)

\(B=2.\frac{100}{101}=\frac{200}{101}\)