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Ta có công thức : \(1+2+3+...+n=\frac{n.\left(n+1\right)}{2}\)
\(\Rightarrow B=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+10}\)
\(=\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+3\right).3}{2}}+...+\frac{1}{\frac{\left(1+10\right)10}{2}}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{10.11}\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{11}\right)=2.\frac{9}{22}=\frac{9}{11}\)
Câu 1:
\(\frac{13}{19}\).(\(\frac{19}{26}\) - \(\frac{71}{43}\)) + \(\frac{71}{43}\).(\(\frac{13}{19}\) - \(\frac{86}{71}\))
= \(\frac{13}{19}\).\(\frac{19}{26}\) - \(\frac{13}{19}\).\(\frac{71}{43}\) + \(\frac{71}{43}\).\(\frac{13}{19}\) - \(\frac{71}{43}\).\(\frac{86}{71}\)
= \(\frac12\) - (\(\frac{13}{19}\).\(\frac{71}{43}\) - \(\frac{13}{19}\).\(\frac{71}{43}\)) - 2
= \(\frac12\) - 0 - 2
= 1/2 - 2
= 1/2 - 4/2
= - 3/2
\(\frac{4.\left(\frac{1}{5}-\frac{1}{71}+\frac{1}{123}\right)}{3.\left(\frac{1}{5}-\frac{1}{71}+\frac{1}{123}\right)}-\frac{\frac{1}{19}+\frac{1}{1315}-\frac{1}{287}}{3.\left(\frac{1}{19}+\frac{1}{1315}-\frac{1}{287}\right)}=\frac{4}{3}-\frac{1}{3}=\frac{3}{3}=1\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}\)
\(A=\frac{1.2.3...99}{2.3.4...100}\)
\(A=\frac{1}{100}\)
\(B=1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{72}\)
\(B=1+1+...+1+\left(\frac{1}{12}+\frac{1}{20}+...+\frac{1}{72}\right)\)
\(B=5.1+\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\right)\)
\(B=5+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}\right)\)
\(B=5+\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(B=5+\frac{2}{9}=\frac{47}{9}\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{99}{100}\)
\(=\frac{1.2.3.4....99}{2.3.4.5...100}\)
\(=\frac{1}{100}\)
Ta có:
\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{70}=\left[1+\frac{1}{70}\right]+\left[\frac{1}{2}+\frac{1}{69}\right]+\left[\frac{1}{3}+\frac{1}{68}\right]+...+\left[\frac{1}{35}+\frac{1}{36}\right]\)
\(=\frac{71}{1.70}+\frac{71}{2.69}+\frac{71}{3.68}+...+\frac{71}{35.36}\)
\(=71\left[\frac{1}{1.70}+\frac{1}{2.69}+\frac{1}{3.68}+...+\frac{1}{35.36}\right]⋮71\)
=> \(A=1\times2\times3\times4\times...\times70\times\left[1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{70}\right]⋮71\)=> ĐPCM
AI THẤY ĐÚNG NHỚ ỦNG HỘ NHA
Xét \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{70}=\left(1+\frac{1}{70}\right)+\left(\frac{1}{2}+\frac{1}{69}\right)+...+\left(\frac{1}{35}+\frac{1}{36}\right)\)
\(=\frac{71}{1.70}+\frac{71}{2.69}+...+\frac{71}{35.36}=71\left(\frac{1}{1.70}+\frac{1}{2.69}+...+\frac{1}{35.36}\right)\)
=>\(A=1.2.3.4...71.\left(\frac{1}{1.70}+\frac{1}{2.69}+...+\frac{1}{35.36}\right)⋮71\)
Vậy A chia hết cho 71
1)
a) \(\frac{9^{14}.25^5.8^7}{18^{12}.625^3.24^3}=\frac{3^{28}.5^{10}.2^{21}}{2^{21}.3^{24}.5^{12}.3^3.2^9}=\frac{3}{5^2}=\frac{3}{25}\)
Bài 2:
\(\frac{abab}{cdcd}=\frac{ab.101}{cd.101}=\frac{ab}{cd};\frac{ababab}{cdcdcd}=\frac{ab.10101}{cd.10101}=\frac{ab}{cd}\)
Vậy \(\frac{ab}{cd}=\frac{abab}{cdcd}=\frac{ababab}{cdcdcd}\)
mk chỉ cần nhìn sơ qua là biết có câu dễ sao bn ko tự nghĩ đi hơi dễ rồi trừ khi bn đố tôi chục câu tiếng anh vật lí văn
\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\)
\(A=1+1+...+1-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)
\(A=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=9-\left(1-\frac{1}{10}\right)=9-1+\frac{1}{10}=8\frac{1}{10}\)