a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\)

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

\(=\frac{1}{2}.\left(1-\frac{1}{7}\right)\)

\(=\frac{1}{2}.\frac{6}{7}\)

\(=\frac{3}{7}\)

b)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009.2011}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)

\(=\frac{1}{2}.\frac{2010}{2011}\)

\(=\frac{1005}{2011}\)

\(C=\left(1+\frac{1}{1.3}\right)\)\(.\left(1+\frac{1}{2.4}\right)\)\(.\left(1+\frac{1}{3.5}\right)\)\(.\left(1+\frac{1}{2014.2016}\right)\)

   \(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{2015^2}{2014.2016}\)

   \(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2015.2015}{2014.2016}\)

   \(=\frac{\left(2.3.4...2015\right).\left(2.3.4...2015\right)}{\left(1.2.3...2014\right).\left(3.4.5...2016\right)}\)

   \(=\frac{2015.2}{2016}\)

    \(=...\)(tự tinhs)

\(C=\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}.\dfrac{25}{4.6}....\dfrac{9801}{9800}=\)

\(=\dfrac{2^2.3^2.4^2.5^2.....99^2}{1.2.3^2.4^2.5^2....98^2.99.100}=\dfrac{2.99}{100}=\dfrac{198}{100}=1,98\)

a) Đặt B= 1/1.3 + 1/3.5 + 1/5.7 + .....+ 1/19.21

Ta có: 2B= 2/1.3 + 2/3.5 + 2/5.7 + ....+ 2/19.21

= 1- 1/3 + 1/3-1/5 + 1/5-1/7 +....+ 1/19-1/21

= 1-1/21 = 20/21

=> B= 20/21 : 2 => B= 10/21

b) Như trên, ta có: 2A= 1- (1/2n + 1) => A=( 1-1/2n+1).1/2

=> A= 1/2- 1/2n+1

=> A< 1/2 ( đpcm )

Em xem lại đề câu B nhé\(B=\dfrac{3}{2}+\dfrac{3}{6}+\dfrac{3}{12}+\dfrac{3}{20}+...+\dfrac{3}{\left(n-1\right).n}\\ =3.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{\left(n-1\right).n}\right)\\ =3.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\right)=3.\left(1-\dfrac{1}{n}\right)=3.\dfrac{n-1}{n}=3-\dfrac{3}{n}.\)

\(C=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{30.32}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{30}-\dfrac{1}{32}\\ =1-\dfrac{1}{32}=\dfrac{31}{32}.\)

\(D=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{n+1}-\dfrac{1}{n+3}\right)\\ =\dfrac{1}{2}.\left(1-\dfrac{1}{n+3}\right)=\dfrac{1}{2}.\dfrac{n+2}{n+3}.\)

a. \(\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{3-1}{3}=\dfrac{2}{3}\); \(\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5-3}{15}=\dfrac{2}{15}\)

b. Ta có \(VP=\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{2}{3}\) mà \(VP=\dfrac{2}{3}\) \(\Rightarrow VT=VP\)

Ta có \(VP=\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{2}{15}\) mà \(VP=\dfrac{2}{3.5}=\dfrac{2}{15}\) \(\Rightarrow VT=VP\)

c. \(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{97.99}+\dfrac{2}{99.101}\)

\(=2\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{97.99}+\dfrac{1}{99.101}\right)\)

\(=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=2\left(1-\dfrac{1}{101}\right)\) \(=\dfrac{200}{101}\)

a: \(\dfrac{1}{1}-\dfrac{1}{3}=1-\dfrac{1}{3}=\dfrac{2}{3}\)

\(\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{2}{15}\)

b: \(\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{3}{3}-\dfrac{1}{3}=\dfrac{2}{3}\)

\(\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{15}-\dfrac{3}{15}=\dfrac{2}{15}\)

c: Ta có: \(A=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=\dfrac{100}{101}\)

P = 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/49.51

P = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/49 - 1/51

P = 1 - 1/51

P = 50/51

Q = 1/1.3 + 1/3.5 + ... + 1/19.21

Q = 1/2 .(2/1.3 + 2/3.5 + ... + 2/19.21)

Q = 1/2.(1 - 1/3 + 1/3 - 1/5 + ... + 1/19 - 1/21)

Q = 1/2 . (1 - 1/21)

Q = 1/2. 20/21

Q = 10/21

Ủng hộ mk nha ^_-

\(P=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)

\(P=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)

\(P=1-\frac{1}{51}\)

\(P=\frac{50}{51}\)

\(Q=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}\)

\(Q=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{19.21}\right)\)

\(Q=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\right)\)

\(Q=\frac{1}{2}.\left(1-\frac{1}{21}\right)\)

\(Q=\frac{1}{2}.\frac{20}{21}\)

\(Q=\frac{10}{21}\)

\(P=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)

\(=1-\frac{1}{51}\)

\(=\frac{50}{51}\)

\(Q=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}\)

\(2Q=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{19.21}\)

\(2Q=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\)

\(2Q=1-\frac{1}{21}\)

\(2Q=\frac{20}{21}\)

\(Q=\frac{20}{21}:2\)

\(Q=\frac{10}{21}\)

còn cần không bạn, mk làm cho