Ta có :

\(A=\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+...+\frac{1}{101\cdot400}\)

\(\Rightarrow299A=\frac{299}{1\cdot300}+\frac{299}{2\cdot301}+...+\frac{299}{101\cdot400}\)

\(\Rightarrow299A=1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}=C\)

\(\Rightarrow A=\frac{C}{299}\)

Lại có :

\(B=\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+\frac{1}{3\cdot104}+...+\frac{1}{299\cdot400}\)

\(\Rightarrow101B=\frac{101}{1\cdot102}+\frac{101}{2\cdot103}+...+\frac{101}{299\cdot400}\)

\(\Rightarrow101B=1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+...+\frac{1}{299}-\frac{1}{400}\)

\(\Rightarrow101B=\left(1+\frac{1}{2}+...+\frac{1}{299}\right)-\left(\frac{1}{102}+\frac{1}{103}+...+\frac{1}{400}\right)\)

\(\Rightarrow B=\frac{C}{101}\)

\(\Rightarrow\frac{A}{B}=\frac{101}{299}\)

299A=\(\frac{299}{1\cdot300}+\frac{299}{2\cdot301}+...+\frac{299}{101\cdot400}\)

299A=\(1-\frac{1}{300}+\frac{1}{300}-\frac{1}{301}-...-\frac{1}{101}+\frac{1}{101}-\frac{1}{400}\)

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

299A=\(\frac{399}{400}\)

A=\(\frac{399}{400}:299\)

A=\(\frac{119310}{400}\)

tương tự tính câu B

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

\(\Rightarrow A=\frac{1}{399}.\left(\frac{299}{1.300}+\frac{299}{2.301}+\frac{299}{3.302}+...+\frac{299}{101.400}\right)\)

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

\(\Rightarrow A=\frac{1}{299}.\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+..+\frac{1}{401}\right)\right]\)

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

\(\Rightarrow B=\frac{1}{101}.\left(1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+...+\frac{1}{299}-\frac{1}{400}\right)\)

\(\Rightarrow B=\frac{1}{101}.\left[\left(1+\frac{1}{2}+...+\frac{1}{299}\right)-\left(\frac{1}{102}+\frac{1}{203}+...+\frac{1}{400}\right)\right]\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{299}.\left[\left(1+\frac{1}{2}+..+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]}{\frac{1}{101}.\left[\left(1+\frac{1}{2}+....+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]}\)

               \(=\frac{1}{299}:\frac{1}{101}=\frac{101}{299}\)

mk chỉ nghĩ đc bài 2 thui mong bn thông cảm.

2) Ta có\(\frac{a}{b}+\frac{b}{a}=\frac{a.a+b.b}{b.a}=\frac{a+b}{1}\)

Mà theo đề bài a,b\(\inℕ^∗\)

=> \(a,b\ge1\Rightarrow a+b\ge2\Rightarrow\frac{a+b}{1}\ge2\)

Thấy đúng thì tk nha