\(A=\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\)

\(A=\frac{1}{299}\cdot\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+\frac{1}{3}-\frac{1}{302}+....+\frac{1}{101}-\frac{1}{400}\right)\)

\(A=\frac{1}{299}\cdot\left(1-\frac{1}{400}\right)\)

\(A=\frac{1}{299}\cdot\frac{399}{400}=\frac{399}{119600}\)

con lau moi tra loi nha ban

trang 21  sách nâng cao và phát triển của vũ hữu bình bn nhé giả có ở trang 86 í

mk ko có sách đấy

bn gửi cho mk đi

mk cảm ơn bạn

Ta có: a=1/1.300+1/2.301+...+1/101.400

⇒ a= 1/299.(299/1.300+299/2.301+...+299/101.400)

⇒ a= 1/299. ( 1+1/300+1/2-1/301+....+1/101-1/400)

⇒ a= 1/299.|(1+1/2+....+1/101)-(1/300+1/301+....+1/400)|

Ta có: b=1/1.102+1/2.103+..+1/299.400

⇒ b= 1/101.(101/1.102+101/2.103+..+101/299.400)

⇒ 1/101.|(1-1/102+1/2-1/102+......+1/299-1/400)|

⇒ b= 1/101 .|(1+1/2+....+1/299) - (1/102+1/103+....+1/400)|

⇒ b= |(1+1/2+....+1/299)- (1/300+1/301+....+1/400)|

⇒a=1/299.|(1+1/2+....+1/101)-(1/300+1/301+....+1/400)|

phần

b=1/101.|(1+1/2+....+1/101)-(1/300+1/301+....+1/400)| 

⇒a/b=1/299:1/101

⇒a/b=101/299.

Ta chú ý đẳng thức \(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\)(Chứng minh rất dễ, bạn quy đồng lên là được nha)

\(A=\frac{1}{1.300}+\frac{1}{2.301}+...+\frac{1}{101.400}\)

\(\Rightarrow299A=\frac{299}{1.300}+\frac{299}{2.301}+\frac{299}{3.302}+...+\frac{299}{101+400}\)

\(=1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+\frac{1}{3}-\frac{1}{302}+...+\frac{1}{101}-\frac{1}{400}\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\)

Đặt \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}=X,\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}=Y\)

\(\Rightarrow A=\frac{X-Y}{299}\)

\(B=\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}\)

\(\Rightarrow101B=\frac{101}{1.102}+\frac{101}{2.103}+\frac{101}{3.104}+...+\frac{101}{299.400}\)

\(=1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+\frac{1}{3}-\frac{1}{104}+...+\frac{1}{102}-\frac{1}{203}+\frac{1}{103}-\frac{1}{204}+...\)

\(\frac{1}{198}-\frac{1}{299}+\frac{1}{199}-\frac{1}{300}+\frac{1}{200}-\frac{1}{301}+...+\frac{1}{299}-\frac{1}{400}\)

\(=\left(1+...+\frac{1}{101}\right)-\left(\frac{1}{300}+...+\frac{1}{400}\right)+\left(\frac{1}{102}-\frac{1}{102}\right)+\left(\frac{1}{103}-\frac{1}{103}\right)+...+\left(\frac{1}{299}-\frac{1}{299}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)=X-Y\)

\(\Rightarrow B=\frac{X-Y}{101}>\frac{X-Y}{299}=A\)

Vậy \(B>A\)