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Câu hỏi của Nguyễn Kim Chi - Toán lớp 6 | Học trực tuyến
quá dễ dàng
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 :
\(\frac{A}{B}=\frac{\left(\frac{1}{400}\right)}{\left(\frac{1}{101}\right)}=\frac{101}{400}\)
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 \(\frac{A}{B}=\frac{1}{200}\)
Ta có :
\(A=\frac{101}{1}+\frac{100}{2}+\frac{99}{3}+...+\frac{1}{101}\)
\(A=\left(101-1-...-1\right)+\left(\frac{100}{2}+1\right)+\left(\frac{99}{3}+1\right)+...+\left(\frac{1}{101}+1\right)\)
\(A=\frac{102}{102}+\frac{102}{2}+\frac{102}{3}+...+\frac{102}{101}\)
\(A=102\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}+\frac{1}{102}\right)\)
\(\Rightarrow\)\(\frac{A}{B}=\frac{102\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{102}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{102}}=\frac{102}{1}=102\)
Vậy \(\frac{A}{B}=102\)
Chúc bạn học tốt ~
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=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}+\frac{1}{101.102}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{102}\)
\(A=1-\frac{1}{102}\)
\(A=\frac{101}{102}\)
\(Tacocongthuc:\frac{1}{n}-\frac{1}{n+k}=\frac{k}{n\left(n+k\right)}\)
\(A=\frac{1}{1.21}+\frac{1}{2.22}+...+\frac{1}{80.100}\Rightarrow20A=\frac{20}{1.21}+\frac{20}{2.22}+.....+\frac{20}{80.100}=1-\frac{1}{21}+\frac{1}{2}-\frac{1}{22}+.....+\frac{1}{80}-\frac{1}{100}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}-\frac{1}{81}-\frac{1}{82}-....-\frac{1}{100}\)
\(B=\frac{1}{1.81}+\frac{1}{2.82}+.....+\frac{1}{20.100}\Rightarrow80B=\frac{80}{1.81}+\frac{80}{2.82}+....+\frac{80}{20.100}=1-\frac{1}{81}+\frac{1}{2}-\frac{1}{82}-....+\frac{1}{20}-\frac{1}{100}=1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{20}-\frac{1}{81}-\frac{1}{82}-.....-\frac{1}{100}\)
\(\Rightarrow20A=80B\Leftrightarrow A=4B\Rightarrow\frac{A}{B}=4\)
sao 1-1/21+1/2-1/22+1/3-1/23+...+1/80-1/100 lại bằng 1+1/2+1/3+...+1/20-1/81-1/82-...-1/100 đúng ra phải là 1+1/2+1/3+...+1/80-1/21-1/22-1/23-...-1/100