\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}\)

=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\)

=\(1-\dfrac{1}{5}\)

=\(\dfrac{4}{5}\)

1/1 - 1/2+ 1/2 - 1/3 + 1/3 - 1/4+ 1/4 - 1/5

 = 1/1 - 1/5

= 4/5

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}=1-\dfrac{1}{5}=\dfrac{4}{5}\)

Nhận xét thấy:

\(\dfrac{1}{1.2}\)= 1-\(\dfrac{1}{2}\); \(\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\);...

Ta có

A= 1-\(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)

A= 1- \(\dfrac{1}{6}\)

A= \(\dfrac{5}{6}\)

Vậy A= \(\dfrac{5}{6}\)

CAU NAY RAT DE NHA BAN

A=\(\dfrac{1}{1}\)-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)

A=1-\(\dfrac{1}{6}\)

=>A=\(\dfrac{5}{6}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)

\(A=1-\dfrac{1}{6}\)

\(A=\dfrac{5}{6}\)

a: \(=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{18\cdot19}-\dfrac{1}{19\cdot20}\)

=1/2-1/380

=179/380

b: \(=\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+...+\dfrac{1}{21\cdot23}-\dfrac{1}{23\cdot25}\)

\(=\dfrac{1}{3}-\dfrac{1}{575}=\dfrac{572}{1725}\)

c: \(=1+\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}-\dfrac{1}{20}-\dfrac{1}{21}\)

=1-1/21

=20/21

d: \(=\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdot...\cdot\left(1-\dfrac{1}{121}\right)\)

\(=\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{10}{11}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{12}{11}\)

\(=\dfrac{2}{11}\cdot\dfrac{12}{2}=\dfrac{12}{11}\)

Ta có:

\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}=\dfrac{9}{10}\)

Bạn có ghi thiếu đề không vậy?

d, `3,15+2,4=5,55`

e, \(\dfrac{5}{7}.\dfrac{2}{11}+\dfrac{5}{7}.\dfrac{9}{11}=\dfrac{5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{5}{7}.\dfrac{11}{11}=\dfrac{5}{7}.1=\dfrac{5}{7}\)

f, `1,25.3,6+3,6.8,75=3,6(1,25+8,75)=3,6.10=36`

\(h,\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}\\ =\dfrac{99}{100}\)

\(e\dfrac{5}{7}\times\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{5}{7}\times1=\dfrac{5}{7}\)

\(f3.6\times\left(1.25+8.75\right)=3.6\times10=36\)

A= \(\dfrac{1}{1.2}\)+ \(\dfrac{1}{2.3}\)+ \(\dfrac{1}{3.4}\)+ \(\dfrac{1}{4.5}\)+ \(\dfrac{1}{5.6}\)

= 1-\(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\)- \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\)- \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)- \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)- \(\dfrac{1}{6}\)

= 1 - \(\dfrac{1}{6}\)= \(\dfrac{5}{6}\)

mk chỉ bt làm câu 1 thôi ak

mong bn thông cảm

\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}=\dfrac{1}{3}\) .-.

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)

\(=\dfrac{1}{2}-\dfrac{1}{6}=\dfrac{3-1}{6}=\dfrac{2}{6}=\dfrac{1}{3}\)

1/3

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{149.150}\)

\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{149}-\dfrac{1}{150}\)

\(A=\dfrac{1}{1}-\dfrac{1}{150}\)

\(A=\dfrac{150}{150}-\dfrac{1}{150}\)

\(A=\dfrac{149}{150}\)