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Ta có:
\(A=\frac{1}{1.300}+\frac{1}{2.301}+...+\frac{1}{101.400}\)
\(\Rightarrow A=\frac{1}{299}.\left(\frac{299}{1.300}+\frac{299}{2.301}+...+\frac{299}{101.400}\right)\)
\(\Rightarrow A=\frac{1}{299}.\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}\right)\)
\(\Rightarrow A=\frac{1}{299}.\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]\)
Lại có:
\(B=\frac{1}{1.102}+\frac{1}{2.103}+...+\frac{1}{299.400}\)
\(\Rightarrow B=\frac{1}{101}.\left(\frac{101}{1.102}+\frac{101}{2.103}+...+\frac{101}{299.400}\right)\)
\(\Rightarrow B=\frac{1}{101}.\left(1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+...+\frac{1}{299}-\frac{1}{400}\right)\)
\(\Rightarrow B=\frac{1}{101}.\left[\left(1+\frac{1}{2}+...+\frac{1}{299}\right)-\left(\frac{1}{102}+\frac{1}{103}+...+\frac{1}{400}\right)\right]\)
\(\Rightarrow B=\frac{1}{101}.\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{299}.\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]}{\frac{1}{101}.\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]}\)
\(\Rightarrow\frac{A}{B}=\frac{1}{299}:\frac{1}{101}\)
\(\Rightarrow\frac{A}{B}=\frac{101}{299}.\)
Vậy \(\frac{A}{B}=\frac{101}{299}.\)
Chúc bạn học tốt!
\(A=\frac{1}{299}.\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...........+\frac{1}{101}-\frac{1}{400}\right)\)
\(=\frac{1}{299}.\left(1+\frac{1}{2}+........+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-....-\frac{1}{400}\right)\)
\(B=\frac{1}{101}.\left(1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+........+\frac{1}{299}-\frac{1}{400}\right)\)
\(=\frac{1}{101}.\left(1+\frac{1}{2}+.......+\frac{1}{299}-\frac{1}{102}-\frac{1}{103}-............-\frac{1}{400}\right)\)
\(=\frac{1}{101}\left(1+\frac{1}{2}+......+\frac{1}{102}+\frac{1}{103}+.....+\frac{1}{299}-\frac{1}{102}-.....-\frac{1}{300}-....-\frac{1}{400}\right)\)
\(=\frac{1}{101}\left(1+\frac{1}{2}+........+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-...-\frac{1}{400}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{299}}{\frac{1}{101}}=\frac{101}{299}\)
299A=(1+1/2+1/3+...+1/101)-(1/300+1/301+...+1/400)=C
101B=(1+1/2+1/3+...+1/299)-(1/102+1/103+..+1/400)=D=C
=>A/B=C/299.101/C=101/299
\(A=\frac{\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{101\cdot400}}{\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+\frac{1}{3\cdot104}+...+\frac{1}{299\cdot400}}\)
\(A=\frac{\frac{1}{299}\left(\frac{299}{1\cdot300}+\frac{299}{2\cdot301}+\frac{299}{3\cdot302}+...+\frac{299}{101\cdot400}\right)}{\frac{1}{101}\left(\frac{101}{1\cdot102}+\frac{101}{2\cdot103}+\frac{101}{3\cdot104}+...+\frac{299}{299\cdot400}\right)}\)
\(A=\frac{\frac{1}{299}\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+\frac{1}{3}-\frac{1}{302}+...+\frac{1}{101}-\frac{1}{400}\right)}{\frac{1}{101}\left(1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+\frac{1}{3}-\frac{1}{104}+...+\frac{1}{299}-\frac{1}{400}\right)}\)
S= 1/199 + 2/198 + ... + 198/2 + 199/1
S= (1/199 + 1) + (2/198 + 1)+ ... + (198/2 + 1) +1
S= 200/200 + 200/199 + 200/198 + ... + 200/2
S= 200.(1/200 + 1/199 + ... + 1/2)
Suy ra , B=(1/2 + 1/3 + ... +1/200) : 200.(1/2 + 1/3 + ... + 1/200)
B=1 : 200 = 1/200
đợi 1 năm nữa rùi mk giải cho!
Ta có:
101A-299B=0
suy ra:101A=299B
suy ra: \(\frac{A}{B}\)=\(\frac{101}{299}\)
Vậy A/B Không phải số nguyên