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a) \(A=\frac{1}{5}+\frac{1}{45}+\frac{1}{117}+\frac{1}{221}+\frac{1}{357}+\frac{1}{525}\)
\(\Rightarrow A=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{21.25}\)
\(\Rightarrow4A=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{21.25}\)
\(4A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{21}-\frac{1}{25}\)
\(4A=\frac{1}{1}-\frac{1}{25}=\frac{24}{25}\)
\(\Rightarrow A=\frac{24}{25}\div4=\frac{6}{25}
a: A>1/150*50+1/200*50=1/3+1/4=7/12
b: A>7/12
7/12>5/8
=>A>5/8
Ta có: \(C=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)
\(=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{120}\right)+\left(\dfrac{1}{121}+\dfrac{1}{122}+\dfrac{1}{123}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+\dfrac{1}{153}+...+\dfrac{1}{180}\right)+\left(\dfrac{1}{181}+\dfrac{1}{182}+\dfrac{1}{183}+...+\dfrac{1}{200}\right)\)
\(\Leftrightarrow C>20\cdot\dfrac{1}{120}+30\cdot\dfrac{1}{150}+30\cdot\dfrac{1}{180}+20\cdot\dfrac{1}{200}\)
\(\Leftrightarrow C>\dfrac{1}{6}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{10}=\dfrac{19}{30}=\dfrac{76}{120}\)
\(\Leftrightarrow C>\dfrac{75}{120}=\dfrac{5}{8}\)
hay \(C>\dfrac{5}{8}\)(đpcm)