Đề hình như bị sai ban ơi sửa lại

\(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{92.95}\)

\(A=3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)

\(A=3.\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)

\(A=\dfrac{1}{2}-\dfrac{1}{95}\)

\(A=\dfrac{93}{190}\)

\(B=\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{92.95}\)

\(3B=2\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)

\(3B=2.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)

\(3B=2\left(\dfrac{1}{2}-\dfrac{1}{95}\right)\)

\(3B=2.\dfrac{93}{190}\)

\(3B=\dfrac{93}{95}\)

\(\Rightarrow B=\dfrac{31}{95}\)

Chịu r

=1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14

=1/2-1/14

=7/14-1/14=6/14=3/7

\(A=\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{92.95}+\dfrac{3}{95.98}\right)\\ A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{98}\right)\\ A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{98}\right)\\ A=\dfrac{1}{3}.\dfrac{24}{49}=\dfrac{8}{49}\)

\(A=\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+..+\dfrac{3}{92.95}+\dfrac{3}{95.98}\right)\)

\(A=\dfrac{3}{2.5.3}+\dfrac{3}{5.8.3}+\dfrac{3}{8.11.3}+..+\dfrac{3}{92.95.3}+\dfrac{3}{95.98.3}\)

\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+..+\dfrac{1}{92.95}+\dfrac{1}{95.98}\)

\(A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+..+\dfrac{1}{92}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{98}\)

\(A=\dfrac{1}{2}-\dfrac{1}{98}=\dfrac{49}{98}-\dfrac{1}{98}=\dfrac{48}{98}=\dfrac{24}{49}\)

1) a) A=\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(=\frac{1}{3}-\frac{1}{8}=\frac{5}{24}\)

c) C=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

\(C=1-\frac{1}{101}\)

\(C=\frac{100}{101}\)

d) Sửa đề: thay \(\frac{3}{92.98}\)=\(\frac{3}{92.95}\)

\(D=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{92}-\frac{1}{95}\)

\(D=\frac{1}{2}-\frac{1}{95}\)

\(D=\frac{95-2}{190}=\frac{93}{190}\)

Các bài trên áp dụng theo tính chất: \(\frac{a}{b\left(b+a\right)}\frac{1}{b}-\frac{1}{b+a}\)

\(S=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{98.101}\)

\(S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\)

\(S=\frac{1}{2}-\frac{1}{101}\)

\(S=\frac{99}{202}\)

Bạn cho mình thêm 1/3 ở dòng 2 của bài làm mình nhé

Kết quả ra là : \(\frac{33}{202}\)

\(\frac{3}{2.5}\)+ \(\frac{3}{5.8}\)+ ...... + \(\frac{3}{92.95}\)= 3 . ( \(\frac{1}{2.5}\)+ \(\frac{1}{5.8}\)+ .... + \(\frac{1}{92.95}\))

= 3 . \(\frac{1}{3}\). ( \(\frac{1}{2.5}\)+ \(\frac{1}{5.8}\)+  ..... + \(\frac{1}{92.95}\))

= 3. \(\frac{1}{3}\). ( \(\frac{1}{2}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{8}\)+ ....... + \(\frac{1}{92}\)- \(\frac{1}{95}\))

= 1 .( \(\frac{1}{2}\)- \(\frac{1}{95}\)) = \(\frac{93}{190}\)

Thấy hay thì cho mình một k nhé!!!

3/ 2.5 + 3/ 5.8 + 3/ 8.11+ ...+ 3/ 92.95

=1/2-1/5+1/5-1/8+1/8-1/11+........+1/92-1/95

=1/2-1/95

=31/60

3/ 2.5 + 3/ 5.8 + 3/ 8.11+ ...+ 3/ 92.95

=1/2-1/5+1/5-1/8+1/8-1/11+........+1/92-1/95

=1/2-1/95

=31/60