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Hỏi Làm Gì

Yuuki Asuna

Chủ đề:

Violympic toán 6

Câu hỏi:

\(\left(\dfrac{1}{1.101}+\dfrac{1}{2.102}+...+\dfrac{1}{10.110}\right).x=\dfrac{1}{1.11}+\dfrac{1}{2.12}+...+\dfrac{1}{100.110} \)
\(\left(\dfrac{1}{100}.\dfrac{100}{1.101}+\dfrac{1}{100}.\dfrac{100}{2.102}+...+\dfrac{1}{100}.\dfrac{100}{10.110}\right).x=\dfrac{1}{10}.\dfrac{10}{1.11}+\dfrac{1}{10}.\dfrac{10}{2.12}+...+\dfrac{1}{10}.\dfrac{10}{100.110}\)
\(\dfrac{1}{100}.\left(\dfrac{1}{1}-\dfrac{1}{101}+\dfrac{1}{2}-\dfrac{1}{102}+...+\dfrac{1}{10}-\dfrac{1}{110}\right).x=\dfrac{1}{10}.\left(\dfrac{1}{1}-\dfrac{1}{11}+\dfrac{1}{2}-\dfrac{1}{12}+...+\dfrac{1}{100}-\dfrac{1}{110}\right)\)
\(\dfrac{1}{100}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{10}\right)-\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{110}\right)\right]\).\(x=\dfrac{1}{10}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{100}\right)-\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{110}\right)\right]\)
\(\dfrac{1}{100}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{10}\right)-\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{110}\right)\right].x=\dfrac{1}{10}\left\{\left(1+\dfrac{1}{2}+...+\dfrac{1}{10}\right)+\left[\left(\dfrac{1}{11}+...+\dfrac{1}{100}\right)-\left(\dfrac{1}{11}+...+\dfrac{1}{100}\right)\right]-\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{110}\right)\right\}\)
\(\dfrac{1}{100}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{10}\right)-\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{110}\right)\right].x=\dfrac{1}{10}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{10}\right)+0-\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{110}\right)\right]\)
\(\Rightarrow\dfrac{1}{100}.x=\dfrac{1}{10}\)
\(x=\dfrac{1}{10}:\dfrac{1}{100}\)
\(x=\dfrac{1}{10}.\dfrac{100}{1}\)
\(x=10\)