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A= 1/1-1/2+1/2-1/3+1/4-1/5+...+1/101-1/102
A=1-1/102=102/102-1/102=101/102
ý b thì chờ mình tí tìm cách lập luận đã nhé
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}+\frac{1}{101.102}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{102}\)
\(A=1-\frac{1}{102}\)
\(A=\frac{101}{102}\)
\(A=\frac{1}{2.2}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(A=\frac{1}{4}+\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(A=\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
( gạch bỏ các phân số giống nhau)
\(A=\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(A=\frac{1}{4}+\frac{2}{9}\)
\(A=\frac{17}{36}\)
phần b, c bn lm tương tự như phần a nha
Ta có :
\(A=\dfrac{1}{1.300}+\dfrac{1}{2.301}+\dfrac{1}{3.302}+..................+\dfrac{1}{101.400}\)
\(299A=\dfrac{299}{1.300}+\dfrac{299}{2.301}+\dfrac{299}{3.302}+..................+\dfrac{299}{101.400}\)
\(299A=1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+.................+\dfrac{1}{101}-\dfrac{1}{400}\)
\(299A=\left(1+\dfrac{1}{2}+.................+\dfrac{1}{101}\right)-\left(\dfrac{1}{300}+\dfrac{1}{301}+.............+\dfrac{1}{400}\right)=C\)
\(\Rightarrow A=\dfrac{C}{299}\)
Lại có :
\(B=\dfrac{1}{1.102}+\dfrac{1}{2.103}+................+\dfrac{1}{299.400}\)
\(101B=\dfrac{101}{1.102}+\dfrac{101}{2.103}+...............+\dfrac{101}{299.400}\)
\(101B=1-\dfrac{1}{102}+\dfrac{1}{2}-\dfrac{1}{103}+...............+\dfrac{1}{299}-\dfrac{1}{400}\)
\(101B=\left(1+\dfrac{1}{2}+...............+\dfrac{1}{299}\right)-\left(\dfrac{1}{102}+\dfrac{1}{103}+...............+\dfrac{1}{400}\right)=C\)\(\Rightarrow B=\dfrac{C}{101}\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{C}{101}:\dfrac{C}{299}=\dfrac{299}{101}\)
~ Chúc bn học tốt ~
Ta có: \(B=\frac{1}{199}+\frac{2}{198}+...+\frac{199}{1}\)
\(=\frac{200-199}{199}+\frac{200-198}{198}+...+\frac{200-1}{1}\)
\(=\frac{200}{199}-\frac{199}{199}+\frac{200}{198}-\frac{198}{198}+...+\frac{200}{1}-\frac{1}{1}\)
\(=\left(\frac{200}{199}+\frac{200}{198}+...+\frac{200}{1}\right)-\left(\frac{199}{199}+\frac{198}{198}+...+\frac{1}{1}\right)\)
\(=200+200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)-199\)
\(=200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)+\frac{200}{200}\)
\(=200\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)}=\frac{1}{200}\)
Ta có :
\(B=\frac{1}{199}+\frac{2}{198}+....+\frac{198}{2}+\frac{199}{1}\)
\(B=1+\frac{1}{199}+1+\frac{1}{198}+....+1+\frac{198}{2}\)
\(B=\frac{200}{199}+\frac{200}{198}+...+\frac{200}{2}\)
\(B=200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)}=\frac{1}{200}\)
Vậy \(\frac{A}{B}=\frac{1}{200}\)