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Đặt: \(\frac{1}{117}=a,\frac{1}{119}=b\)
Khi đó: \(A=3ab-4a.5.118b-5ab+\frac{8}{39}\)
\(=-2362ab+\frac{8}{39}\)
\(=-2362.\frac{1}{117}.\frac{1}{119}=\frac{38}{1071}\)
\(\frac{1}{20}+\frac{1}{44}+\frac{1}{77}+...+\frac{2}{x\left(x+3\right)}=\frac{101}{770}\)
\(\Rightarrow\)\(\frac{3}{2}.\left(\frac{1}{20}+\frac{1}{44}+\frac{1}{77}+...+\frac{2}{x\left(x+3\right)}\right)=\frac{101}{770}\).
\(\Rightarrow\)\(\frac{3}{40}+\frac{3}{88}+\frac{3}{154}+...+\frac{3}{x\left(x-3\right)}=\frac{303}{1540}\)
\(\Rightarrow\)\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\Rightarrow\)\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x-1}\)
\(\Rightarrow\)\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\)\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\Rightarrow\)\(\frac{1}{x+3}=\frac{308}{1540}-\frac{303}{1540}\)
\(\Rightarrow\)\(\frac{1}{x+3}=\frac{5}{1540}\)
\(\Rightarrow\)\(\frac{1}{x+3}=\)\(\frac{1}{308}\)
\(\Rightarrow\)\(x+3=308\)
\(\Rightarrow\)\(x=308-3\)
\(\Rightarrow\)\(x=305\)
a)Ta đặt A=10+15+...+300
Số số hạng của A là:(300-10):5+1=59(số)
Tổng của A là:(10+300).59:2=9145
=>9145+x=6750
=>x=6750-9145
=>x=-2395
b)\(\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}-\frac{1}{x+1}=\frac{59}{77}\)
<=>\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{6.7}-\frac{1}{x+1}=\frac{59}{77}\)
<=>\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}-\frac{1}{x+1}=\frac{59}{77}\)
<=>\(1-\frac{1}{7}-\frac{1}{x+1}=\frac{59}{77}\)
<=>\(\frac{6}{7}-\frac{1}{x+1}=\frac{56}{77}\)
<=>\(\frac{1}{x+1}=\frac{6}{7}-\frac{56}{77}=\frac{66}{77}-\frac{56}{77}\)
<=>\(\frac{1}{x+1}=\frac{10}{77}\)
<=>10(x+1)=77
<=>10x+10=77
<=>10x=67
<=>x=6,7
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}-1-\frac{1}{2}-...-\frac{1}{1007}\)
\(=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}\) (đpcm)
\(\frac{1}{5}+\frac{1}{20}+\frac{1}{44}+\frac{1}{77}+\frac{1}{119}+\frac{1}{170}+\frac{1}{230}+\frac{1}{299}\)
\(=\frac{1}{1.5}+\frac{1}{5.4}+\frac{1}{4.11}+\frac{1}{11.7}+...+\frac{1}{23.13}\)
\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{4}+...+\frac{1}{23}-\frac{1}{13}\)
\(=1-\frac{1}{13}\)
\(=\frac{12}{13}\)