d) Ta có: \(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{-37}{45}\)

\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{-37}{45}\)

\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{-37}{45}\)

\(\Leftrightarrow x=\dfrac{-37}{45}+\dfrac{1}{45}-\dfrac{1}{5}=\dfrac{-36}{45}-\dfrac{1}{5}=\dfrac{-4}{5}-\dfrac{1}{5}=-1\)

Vậy: x=-1

1)

2/3.5+2/5.7+...+2/11.13+2/13.15+2/1.2+2/2.3+...+2/9.10

=(2/3.5+...2/13.15)+(2/1.2+...+2/9.10)

= (2/3-2/15)+ [2(1-1/10)]

=8/15+9/5

=7/3

2)

11/12+11/12.24+...+11/88.99

=11-1/9

=10/8/9

Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)

A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)

Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)

\(=\dfrac{1}{3}-\dfrac{1}{15}\)

\(=\dfrac{4}{15}\)

Câu 1:

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\)

= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

= \(\dfrac{1}{3}-\dfrac{1}{101}\)

= \(\dfrac{98}{303}\)

Câu 2 làm tương tự ở câu 1 nhé

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+.....+\dfrac{1}{13}-\dfrac{1}{15}\)

(do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với \(a\in N\)*)

\(=\dfrac{1}{3}-\dfrac{1}{15}=\dfrac{4}{15}\)

Chúc bạn học tốt!!!

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)

= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\)

= \(\dfrac{1}{3}-\dfrac{1}{15}\)

= \(\dfrac{4}{15}\)

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\)\(=\dfrac{1}{3}-\dfrac{1}{15}\)

\(=\dfrac{4}{15}\)

Ta có: \(N=\dfrac{5}{3.5}-\dfrac{5}{5.7}-\dfrac{5}{7.9}-\dfrac{5}{9.11}-\dfrac{5}{11.13}-\dfrac{5}{13.15} \)
\(\Rightarrow N=5\left(\dfrac{1}{3.5}-\dfrac{1}{5.7}-\dfrac{1}{7.9}-\dfrac{1}{9.11}-\dfrac{1}{11.13}-\dfrac{1}{13.15}\right)\)
\(\Rightarrow N=\dfrac{5}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{11}-\dfrac{1}{11}+\dfrac{1}{13}-\dfrac{1}{13}+\dfrac{1}{15}\right)\)
\(\Rightarrow N=\dfrac{5}{2}\left(\dfrac{1}{3}+\dfrac{1}{15}\right)=\dfrac{5}{2}\left(\dfrac{5+1}{15}\right)=\dfrac{5}{2}.\dfrac{6}{15}\)
\(\Rightarrow N=\dfrac{3}{3}=1\)

\(=2x+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{11}=-\frac{2016}{2017}\)

\(=2x+\frac{1}{3}-\frac{1}{11}=-\frac{2016}{2017}\)

\(2x+\frac{8}{33}=-\frac{2016}{2017}\)

\(2x=\frac{-2016}{2017}-\frac{8}{33}\)

\(2x=\frac{-2024}{2017}\)

\(x=-\frac{1012}{2017}\)

\(2x+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}=\frac{-2016}{2017}\)

\(2x+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}=\frac{-2016}{2017}\)

\(2x+\frac{1}{3}-\frac{1}{11}=\frac{-2016}{2017}\)

\(2x+\frac{8}{33}=\frac{-2016}{2017}\)

\(2x=\frac{-2016}{2017}-\frac{8}{33}\)

Số dư dài quá. Đến đây bạn tự làm tiếp nhé

\(\text{ }=2x+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{11}=-\frac{2016}{2017}\)

\(=2x+\frac{1}{3}-\frac{1}{11}=-\frac{2016}{2017}\)

\(2x+\frac{8}{33}=-\frac{2016}{2017}\)

\(2x=\frac{-2016}{2017}-\frac{8}{33}\)

\(2x=\frac{-2024}{2017}\)

\(x=-\frac{1012}{2017}\)

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

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

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

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

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

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

\(x=1\)

1)           P = 2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13 + 2/13.15

              P= (1/3-1/5) + (1/5-1/7) + (1/7-1/9) + (1/9-1/11) + (1/11-1/13) + (1/13-1/15)

              P=1/3-1/15= 4/15

2) a/ 0,2:1+3/5+80%

= 2/10:8/5+8/10

= 2/10.5/8+8/10

= 1/8 + 4/5 = 5/40 + 32/40 = 37/40

    b/ 0,5:5/4-2+1/5

= 5/10:5/4-11/5

= 5/10.4/5-11/5

=2/5-11/5 = -9/5