a/ \(A=\dfrac{6}{5.8}+\dfrac{22}{8.19}+\dfrac{24}{19.31}+\dfrac{198}{101.200}\)
\(=2\left(\dfrac{3}{5.8}+\dfrac{11}{8.19}+\dfrac{12}{19.31}+...+\dfrac{99}{101.200}\right)\)
\(=2\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+....+\dfrac{1}{101}-\dfrac{1}{200}\right)\)
\(=2\left(\dfrac{1}{5}-\dfrac{1}{200}\right)\)
\(=\dfrac{39}{100}\)
b/ \(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{100^2}\)
Ta có :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
...........
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{99.100}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Leftrightarrow A< 1-\dfrac{1}{100}< 1\left(đpcm\right)\)
Giải:
a) \(A=\dfrac{6}{5.8}+\dfrac{22}{8.19}+\dfrac{24}{19.31}+\dfrac{140}{31.101}+\dfrac{198}{101.200}\)
\(A=2.\left(\dfrac{3}{5.8}+\dfrac{11}{8.19}+\dfrac{12}{19.31}+\dfrac{70}{31.101}+\dfrac{99}{101.200}\right)\)
\(A=2.\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{101}+\dfrac{1}{101}-\dfrac{1}{200}\right)\)
\(A=2.\left(\dfrac{1}{5}-\dfrac{1}{200}\right)\)
\(A=2.\dfrac{39}{200}\)
\(A=\dfrac{39}{100}\)
b) \(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
Ta có:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\)
\(...\)
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Rightarrow\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(\Rightarrow\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100}-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}\right)\)
\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}...+\dfrac{1}{99}+\dfrac{1}{100}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)
\(\Rightarrow\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{99}+\dfrac{1}{100}\)
Bạn tự lm theo đề bài của bạn nhé vì đề bài chỉ thế này thôi!