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1.
\(A=\frac{1}{199}+\frac{2}{198}+...+\frac{199}{1}\)
cộng 1 vào mỗi phân số trong 198 phân số đầu, trừ phân số cuối đi 198 ta được :
\(A=\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+...+\left(\frac{199}{1}-198\right)\)
\(A=\frac{200}{199}+\frac{200}{198}+...+1\)
\(A=\frac{200}{199}+\frac{200}{198}+...+\frac{200}{200}\)
đưa phân số cuối lên đầu ta được :
\(A=\frac{200}{200}+\frac{200}{199}+\frac{200}{198}+...+\frac{200}{2}\)
\(A=200.\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{200.\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}=200\)
2.
\(A=\frac{1}{1.400}+\frac{1}{2.401}+\frac{1}{3.402}+...+\frac{1}{101.500}\)
\(A=\frac{1}{400}.\left(1-\frac{1}{400}\right)+\frac{1}{400}.\left(\frac{1}{2}-\frac{1}{401}\right)+\frac{1}{400}.\left(\frac{1}{3}-\frac{1}{402}\right)+...+\frac{1}{400}.\left(\frac{1}{101}-\frac{1}{500}\right)\)
\(A=\frac{1}{400}.\left(1-\frac{1}{400}+\frac{1}{2}-\frac{1}{401}+\frac{1}{3}-\frac{1}{402}+...+\frac{1}{101}-\frac{1}{500}\right)\)
\(A=\frac{1}{400}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}-\frac{1}{400}-\frac{1}{401}-\frac{1}{402}-...-\frac{1}{500}\right)\)
\(B=\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{399.500}\)
\(B=\frac{1}{101}.\left(1-\frac{1}{102}\right)+\frac{1}{101}.\left(\frac{1}{2}-\frac{1}{103}\right)+\frac{1}{101}.\left(\frac{1}{3}-\frac{1}{104}\right)+...+\frac{1}{101}.\left(\frac{1}{399}-\frac{1}{500}\right)\)
\(B=\frac{1}{101}.\left(1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+\frac{1}{3}-\frac{1}{104}+...+\frac{1}{399}-\frac{1}{500}\right)\)
\(B=\frac{1}{101}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{399}-\frac{1}{102}-\frac{1}{103}-\frac{1}{104}-...-\frac{1}{500}\right)\)
\(B=\frac{1}{101}.\left(1+\frac{1}{2}+...+\frac{1}{101}+\frac{1}{102}+...+\frac{1}{399}-\frac{1}{102}-...-\frac{1}{399}-\frac{1}{400}-...-\frac{1}{500}\right)\)
\(B=\frac{1}{101}.\left(1+\frac{1}{2}+...+\frac{1}{101}-\frac{1}{400}-...-\frac{1}{500}\right)\)
Ta thấy vế trong ngoặc của hai biểu thức A và B giống nhau, do đó :
\(\frac{A}{B}=\frac{\left(\frac{1}{400}\right)}{\left(\frac{1}{101}\right)}=\frac{101}{400}\)
A=11.300+12.301+13.302+...+1101.400�=11.300+12.301+13.302+...+1101.400
A=1299.(11−1300+12−1301+13−13012+...+1101−1400)�=1299.(11−1300+12−1301+13−13012+...+1101−1400)
A=1299.(11−1400)�=1299.(11−1400)
A=1299.399400�=1299.399400
A=399119600�=399119600
B=11.102+12.103+13.104+...+1299.400�=11.102+12.103+13.104+...+1299.400
B=1101.(11−1102+12−1103+....+1299−1400)�=1101.(11−1102+12−1103+....+1299−1400)
B=1101.(11−1400)�=1101.(11−1400)
B=1101.399400�=1101.399400
B=39940400�=39940400
⇒AB=39911960039940400=101299
A=11.300+12.301+13.302+...+1101.400�=11.300+12.301+13.302+...+1101.400
A=1299.(11−1300+12−1301+13−13012+...+1101−1400)�=1299.(11−1300+12−1301+13−13012+...+1101−1400)
A=1299.(11−1400)�=1299.(11−1400)
A=1299.399400�=1299.399400
A=399119600�=399119600
B=11.102+12.103+13.104+...+1299.400�=11.102+12.103+13.104+...+1299.400
B=1101.(11−1102+12−1103+....+1299−1400)�=1101.(11−1102+12−1103+....+1299−1400)
B=1101.(11−1400)�=1101.(11−1400)
B=1101.399400�=1101.399400
B=39940400�=39940400
⇒AB=39911960039940400=101299
\(A=\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\)
\(A=\frac{1}{299}.\left(\frac{1}{1}-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+\frac{1}{3}-\frac{1}{3012}+...+\frac{1}{101}-\frac{1}{400}\right)\)
\(A=\frac{1}{299}.\left(\frac{1}{1}-\frac{1}{400}\right)\)
\(A=\frac{1}{299}.\frac{399}{400}\)
\(A=\frac{399}{119600}\)
\(B=\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}\)
\(B=\frac{1}{101}.\left(\frac{1}{1}-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+....+\frac{1}{299}-\frac{1}{400}\right)\)
\(B=\frac{1}{101}.\left(\frac{1}{1}-\frac{1}{400}\right)\)
\(B=\frac{1}{101}.\frac{399}{400}\)
\(B=\frac{399}{40400}\)
\(\Rightarrow\frac{A}{B}=\frac{399}{\frac{119600}{\frac{399}{40400}}}=\frac{101}{299}\)