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\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=2.\frac{49}{100}\)
\(=\frac{49}{50}\)
= 2.(1/2.3 + 1/3.4 + ... + 1/99.100)
trong ngoac co cong thuc do, tim hieu di la lam dc
A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
=\(1-\frac{1}{50}\)
Vì \(1-\frac{1}{50}< 1\)nên A < 1
B = \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
=\(\frac{1}{2}-\frac{1}{100}\)
Vì \(\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)nên B < \(\frac{1}{2}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(\Rightarrow A< 1\)
\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=\frac{1}{2}-\frac{1}{100}\)
\(\Rightarrow B< \frac{1}{2}\)
\(X=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(X=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(X=1-\frac{1}{100}=\frac{99}{100}\).
x = 1/1*2 + 1/2*3 +1/3*4 + 1/4*5 + ... + 1/99*100
x = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/99 - 1/100
x = 1- 1/100
x = 99/100
\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{99.100}\)
\(\frac{2}{1}\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)
\(\frac{2}{1}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\frac{2}{1}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(\frac{2}{1}.\frac{49}{100}\)
\(\frac{98}{100}=\frac{49}{50}\)
Đặt A = \(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{99.100}\)
A : 2 = \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
A : 2 = \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
A : 2 = \(\frac{1}{2}-\frac{1}{100}\)
A : 2 = \(\frac{49}{100}\)
A = \(\frac{49}{50}\)
\(\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{99\cdot100}\)
\(=2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=2\cdot\frac{49}{100}\)
\(=\frac{49}{50}\)
=2(\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+...+\(\frac{1}{99.100}\))
=2(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...+\(\frac{1}{99}\)-\(\frac{1}{100}\))
=2(\(\frac{1}{2}\)-\(\frac{1}{100}\))
=2.\(\frac{49}{100}\)
=\(\frac{49}{50}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{99-98}{98.99}+\frac{100-99}{99.100}\)
\(=\frac{2}{1.2}-\frac{1}{1.2}+\frac{3}{2.3}-\frac{2}{2.3}+\frac{4}{3.4}-\frac{3}{3.4}+\frac{5}{4.5}-\frac{4}{4.5}+...+\frac{99}{98.99}-\frac{98}{98.99}+\frac{100}{99.100}-\frac{99}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
A= 1/1-1/2+1/2-1/3+1/4-1/5+...+1/101-1/102
A=1-1/102=102/102-1/102=101/102
ý b thì chờ mình tí tìm cách lập luận đã nhé
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}+\frac{1}{101.102}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{102}\)
\(A=1-\frac{1}{102}\)
\(A=\frac{101}{102}\)
\(A=\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+\frac{2}{4}-\frac{2}{5}+\frac{2}{5}-\frac{2}{6}+.....+\frac{2}{99}-\frac{2}{100}\)
Ta tính các số âm và số dương giống nhau cộng lại có tổng bằng 0
\(\Rightarrow A=\frac{2}{2}-\frac{2}{100}\)
\(A=\frac{100}{100}-\frac{2}{100}=\frac{98}{100}=\frac{49}{50}\)
Đúng 100%
Đúng 100%
Đúng 100%
\(A=\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+....+\frac{2}{99\cdot100}\)
\(A:2=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+....+\frac{1}{99\cdot100}\)
A : 2 = \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)
\(A:2=\frac{1}{2}-\frac{1}{100}\)
\(A:2=\frac{49}{100}\)
A = \(\frac{49}{50}\)