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Ta có:
\(\frac{1}{20.21}+\frac{1}{21.22}+\frac{1}{22.23}+...+\frac{1}{60.61}\)
\(=\frac{1}{20}-\frac{1}{21}+\frac{1}{21}-\frac{1}{22}+\frac{1}{22}-\frac{1}{23}+...+\frac{1}{60}-\frac{1}{61}\)
\(=\frac{1}{2}-\frac{1}{61}=\frac{59}{122}\)
b) \(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{45.49}\)
\(=\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{45.49}\)
\(=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{45}-\frac{1}{49}\)
\(=\frac{1}{5}-\frac{1}{49}=\frac{44}{245}\)
Bn Tấn sai rùi
phần a , câu cuối là \(\frac{1}{20}\)chứ đâu phải \(\frac{1}{2}\)
Đặt \(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{21.22}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{21}-\frac{1}{22}\)
\(A=\frac{1}{5}-\frac{1}{22}\)
\(A=\frac{17}{110}\)
\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)+...+\(\frac{1}{21.22}\)
=\(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)+...+\(\frac{1}{21}\)-\(\frac{1}{22}\)
=\(\frac{1}{5}\)-\(\frac{1}{22}\)
=\(\frac{17}{110}\)
a: \(\left(a-b-c\right)-\left(-c+b+a\right)-\left(a-b\right)\)
\(=a-b-c+c-b-a-a+b\)
\(=-a-b\)
b: \(a\left(b+c\right)-a\left(b+d\right)-\left(1+ac-ad\right)\)
\(=ab+ac-ab-ad-1-ac+ad\)
=-1
a)\(\frac{1}{4}\)(\(\frac{21.22+76.79+81.85}{21.22+76.79+81.85}\))=\(\frac{1}{4}\)
b)\(\frac{20052005}{20062006}\)=\(\frac{20052005:10001}{20062006:10001}\)=\(\frac{2005}{2006}\)