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thực ra nó rất là dễ. giờ mình mới phát hiện ra chứ bữa trước mình làm cách dài lắm
Ta có :
\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)\)
\(=\frac{25}{12}+\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)>\frac{25}{12}\)( đpcm )
Ta có :
\(B=\frac{1}{199}+\frac{2}{198}+...+\frac{198}{2}+\frac{199}{1}\)
\(B=\frac{1}{199}+\frac{2}{198}+...+\frac{198}{2}+199\)
\(B=\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+...+\left(\frac{198}{2}+1\right)+1\)
\(B=\frac{200}{199}+\frac{200}{198}+...+\frac{200}{2}+1\)
\(B=\frac{200}{200}+\frac{200}{199}+\frac{200}{198}+...+\frac{200}{2}\)
\(B=200.\left(\frac{1}{2}+...+\frac{1}{198}+\frac{1}{199}+\frac{1}{200}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+...+\frac{1}{198}+\frac{1}{199}+\frac{1}{200}}{200.\left(\frac{1}{2}+...+\frac{1}{198}+\frac{1}{199}+\frac{1}{200}\right)}=\frac{1}{200}\)
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...........................
\(B=\frac{1}{199}+\frac{2}{198}+...+\frac{199}{1}=\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+...+\left(199-1-...-1\right)\)(198 số 1)
\(=\frac{200}{199}+\frac{200}{198}+...+1=200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}+\frac{1}{200}\right)=200.A\)
\(\Rightarrow\frac{A}{B}=\frac{A}{200A}=\frac{1}{200}\)
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 \(\frac{A}{B}=\frac{1}{200}\)
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 :
\(\frac{A}{B}=\frac{\left(\frac{1}{400}\right)}{\left(\frac{1}{101}\right)}=\frac{101}{400}\)
Tách tổng A thành 4 nhóm
A = ( 1 + \(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\)) + ( \(\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{99}+\frac{1}{100}\right)+\left(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{149}+\frac{1}{150}\right)+\left(\frac{1}{151}+\frac{1}{152}+\frac{1}{153}+...+\frac{1}{199}+\frac{1}{200}\right)\)
A > \(\frac{1}{50}.50+\frac{1}{100}.50+\frac{1}{150}.50+\frac{1}{200}.50\)= \(\left(\frac{1}{50}+\frac{1}{100}+\frac{1}{150}+\frac{1}{200}\right).50=\frac{1}{24}.50=\frac{25}{12}\)
\(\Rightarrow\) A > \(\frac{25}{12}\)