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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)

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

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

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

Nhầm tưởng tính tích :v

Ta có :

\(B=\frac{1}{50}+\frac{1}{51}+...+\frac{1}{99}+\frac{1}{100}< \frac{1}{51}+\frac{1}{51}+...+\frac{1}{51}=50.\frac{1}{51}=\frac{50}{51}< \frac{99}{100}\)

\(\Leftrightarrow A>B\)

~ Rim Ceil ~:Chuyên Quốc Học ở đâu thì ko biết nhưng bài như thế này mak làm sai~

\(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}\)

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

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

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

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

\(=\frac{1}{51}+\frac{1}{52}+....+\frac{1}{99}+\frac{1}{100}\)

\(\Rightarrow A-B=0\)

http://olm.vn/hoi-dap/question/126681.html

Bạn tham khảo nhé

cho 1 cai dung minh cho ban roi ma

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:

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

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

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

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

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

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

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

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

\(\Rightarrow P=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)\(\Rightarrow P=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

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

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

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

Vậy E = 1

cô trang dạy rồi mà

Khó kinh .."

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

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

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

\(\Rightarrow Q=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-2\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)

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

\(\Rightarrow P=1\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(=2013\in Z\)

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