Đặt :
\(A=\dfrac{1}{1.4}+\dfrac{1}{4.7}+.................+\dfrac{1}{97.100}\)
\(3A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+.................+\dfrac{3}{97.100}\)
\(3A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+..........+\dfrac{1}{97}-\dfrac{1}{100}\)
\(3A=1-\dfrac{1}{100}\)
\(3A=\dfrac{99}{100}\)
\(\Rightarrow A=\dfrac{99}{100}:3\)
\(\Rightarrow A=\dfrac{33}{100}\)
~ Chúc bn học tốt ~
Ta có : \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{97.100}\)
= 3( \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{97.100}\) ).\(\dfrac{1}{3}\)
= (\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\)).\(\dfrac{1}{3}\)
= \(\left(\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{100-97}{97.100}\right).\dfrac{1}{3}\) = \(\left(\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{100}{97.100}-\dfrac{97}{97.100}\right).\dfrac{1}{3}\) = \(\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right).\dfrac{1}{3}\) =\(\left(1-\dfrac{1}{100}\right).\dfrac{1}{3}\)
= \(\dfrac{99}{100}.\dfrac{1}{3}\)
= \(\dfrac{99}{100.3}\)
= \(\dfrac{33}{100}\)
Lời giải:
Đặt \(A=\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}\)
\(\Rightarrow A=\dfrac{1.3}{1.4.3}+\dfrac{1.3}{4.7.3}+...+\dfrac{1.3}{97.100.3}\)
\(\Rightarrow A=\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)
\(\Rightarrow A=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(\Rightarrow A=\dfrac{1}{3}.\left(1-\dfrac{1}{100}\right)\)
\(\Rightarrow A=\dfrac{1}{3}.\dfrac{99}{100}\)
\(\Rightarrow A=\dfrac{33}{100}\)
Đặt
B=1/1.4+1/4.7+...+1/97.100
B=1/3.(3/1.4+3/4.7+...+3/97.100)
B=1/3.(1/1-1/4+1/4-1/7+...+1/97-1/100)
B=1/3.(1-1/100)
B=1/3.99/100
B=1.99/3.100
B=1.33.3/3.100
B=33/100
Vậy B= 33/100
Mình có cách ngắn gọn hơn
(1/1-1/100):3=33/100
