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a)
\(A=\left(\frac{1}{9}-\frac{1}{10}\right)-\left(\frac{1}{8}-\frac{1}{9}\right)-....-\left(1-\frac{1}{2}\right)=\frac{1}{9}-\frac{1}{10}-\frac{1}{8}+\frac{1}{9}-....-1+\frac{1}{2}\)
\(A=-\left(\frac{1}{10}+1\right)=-\frac{11}{10}\)
a)\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\\ \Rightarrow A=-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}-\frac{1}{90}\\ \Rightarrow A=-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)Đặt \(B=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{72}+\frac{1}{90}\)
\(\Rightarrow B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(\Rightarrow B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow B=1-\frac{1}{10}=\frac{9}{10}\)
Ta có : \(A=-B\)
\(\Rightarrow A=-\frac{9}{10}\)
mk làm bài 1 thui,bài 2 chỉ qui đồng ms
3a/6 = 3b/4 => 3(a-b)/ (6-4) = 3.4,5/2= 13,5/2 =k
a = 2k=13,5
b = 4k/3 =9
\(A=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}^2-...-\frac{1}{5}\right)\left(2,4.42-21.4,8\right)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)
=> \(A=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}-...-\frac{1}{5}\right).0}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)= 0