\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)

\(=\)\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}\)

\(=\frac{10}{20}-\frac{1}{20}\)

\(=\frac{9}{20}\)

Tk giúp !!

a) Đặt \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}< \frac{1}{2}\)

Vậy A<\(\frac{1}{2}\).

b) Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)

                  \(\frac{1}{3^2}< \frac{1}{2.3}\)

                  \(\frac{1}{4^2}< \frac{1}{3.4}\)

                    ...

                   \(\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(B< 1-\frac{1}{100}< 1\)

Vậy \(B< 1\).

a: =35/17-18/17-9/5+4/5

=1-1=0

b: =-7/19(3/17+8/11-1)

=7/19*18/187=126/3553

c: =26/15-11/15-17/3-6/13

=1-6/13-17/3

=7/13-17/3=-200/39

b: \(27D=3^{14}+3^{17}+...+3^{2024}\)

\(\Leftrightarrow26D=3^{2024}-3^{11}\)

hay \(D=\dfrac{3^{2024}-3^{11}}{26}\)

c: \(25E=-5^4-5^6-...-5^{1002}\)

\(\Leftrightarrow24E=-5^{1002}+5^2\)

hay \(E=\dfrac{-5^{1002}+5^2}{24}\)

a: \(=3\cdot\dfrac{6+14-9}{42}\cdot\dfrac{14}{11}\)

\(=3\cdot\dfrac{11}{42}\cdot\dfrac{14}{11}=1\)

b: \(=\dfrac{19}{5}\cdot\dfrac{11}{10}+\dfrac{9}{5}\cdot\dfrac{7}{3}\)

\(=\dfrac{209}{50}+\dfrac{63}{15}=4.18+4.2=8,38\)

Lời giải chi tiết:

Phương pháp giải:

Tính nhẩm các số rồi điền kết quả vào chỗ trống.

Lời giải chi tiết:

Tính nhẩm:

a; \(\dfrac{9}{4}\) - \(\dfrac{-11}{4}\)

= \(\dfrac{9}{4}\) + \(\dfrac{11}{4}\)

= \(\dfrac{20}{4}\)

= 5 

b; \(\dfrac{7}{8}\) - \(\dfrac{3}{-8}\) - \(\dfrac{1}{8}\)

=  \(\dfrac{7}{8}\) + \(\dfrac{3}{8}\) - \(\dfrac{1}{8}\)

= \(\dfrac{7+3-1}{8}\)

= \(\dfrac{9}{8}\) 

c; \(\dfrac{-5}{21}\) - \(\dfrac{25}{21}\) - \(\dfrac{-1}{21}\)

  = \(\dfrac{-5}{21}\) - \(\dfrac{25}{21}\) + \(\dfrac{1}{21}\)

 =  \(\dfrac{-5-25+1}{21}\)

= \(\dfrac{-29}{21}\)

\(A=\frac{2}{7}+\frac{-3}{8}+\frac{11}{7}+\frac{1}{3}+\frac{1}{7}+\frac{5}{-8}\)

\(A=\left(\frac{2}{7}+\frac{11}{7}+\frac{1}{7}\right)+\left(\frac{-3}{8}+\frac{5}{-8}\right)+\frac{1}{3}\)

\(A=2-1+\frac{1}{3}\)

\(A=\frac{4}{3}\)

\(B=\frac{3}{17}+\frac{-5}{13}+\frac{-18}{35}+\frac{14}{17}+17\)

\(B=\left(\frac{3}{17}+\frac{14}{17}\right)+\frac{-5}{13}+\frac{-18}{35}+17\)

\(B=1+\frac{-5}{13}+\frac{-18}{35}+17\)

\(B=18+\frac{-5}{13}+\frac{-18}{35}\)

\(B=\frac{7781}{455}\)