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Bạn tham khảo nhé

đặt A = 1/1*2 +  1/3*4 + 1/5*6 + ... + 1/99*100

= 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... + 1/99 - 1/100

= (1 + 1/3 + 1/5 + ... + 1/99) - (1/2 + 1/4 + 1/6 + ... + 1/100)

= 1 + 1/2 + 1/3 + ... + 1/100 - 2(1/2 + 1/4 + 1/6 + .... + 1/100)

= 1 + 1/2 + 1/3 + ... + 1/100 - 1 - 1/2 - 13 - ... - 1/50

= 1/51 + 1/52 + 1/53 + ... + 1/100

thay vào ra E = 1

Biến đổi mẫu ta được:

\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

\(\Rightarrow E=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}=1\)

\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{99}-\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

\(\Rightarrow\frac{B}{A}=\frac{2013\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)}{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}=2013\)là số nguyên

\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{97.98}+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+..+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)

\(\Rightarrow\frac{B}{A}=\frac{\frac{2013}{51}+\frac{2013}{52}+..+\frac{2013}{100}}{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}}\)

\(=\frac{2013\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\right)}{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}}\)

\(=2013\in Z\)

Xét vế trái: A\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

=>\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

=>\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

=>\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

=>\(A=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}=VP\)

=>đpcm         (VP là vế phải)

Tính $E=\frac{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+....+\frac{1}{100}}{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+..+\frac{1}{99.100}}$E=151 +152 +153 +....+1100 11.2 +13.4 +15.6 +..+199.100  

Toán lớp 6

Rút gọn mẫu ta được:

\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+....+\frac{1}{100}\)

Vì tử và mẫu bằng nhau nên biểu thức bằng 1

Bạn muốn biết cách rút gọn mẫu thì gửi tin nhắn cho mình

+) \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(A=\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+...+\frac{1}{9900}\)

\(A=\left(\frac{1}{2}+\frac{1}{12}\right)+\left(\frac{1}{30}+...+\frac{1}{9900}\right)>\frac{1}{2}+\frac{1}{12}.\)

\(\Rightarrow A>\frac{1}{2}+\frac{1}{12}\)

\(\Rightarrow A>\frac{7}{12}\left(1\right).\)

+) \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\left(1-\frac{1}{2}+\frac{1}{3}\right)-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{5}{6}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}< \frac{5}{6}\)

\(\Rightarrow A< \frac{5}{6}\left(2\right).\)

Từ \(\left(1\right)và\left(2\right)\Rightarrow\frac{7}{12}< A< \frac{5}{6}\left(đpcm\right).\)

Chúc bạn học tốt!

bài này lớp 6 mik làm rùi

Ta có:

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\) 

\(A=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)

Ta có \(\frac{B}{A}=2011\)

bạn ơi mình vẫn chưa hiểu lắm