Bài 2.6:
\(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-.....-\dfrac{1}{2.1}\)
\(C=-\left(\dfrac{-1}{100}+\dfrac{1}{100.99}+......+\dfrac{1}{2.1}\right)\)
\(C=-\left(\dfrac{-1}{100}+\dfrac{1}{100}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{98}+....+\dfrac{1}{2}-\dfrac{1}{1}\right)\)
\(C=-1\)
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2.4:
\(A=\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\left(\dfrac{1}{3}-\dfrac{2}{9}\right)-\dfrac{3}{4}+\left(\dfrac{3}{5}+\dfrac{1}{15}\right)+\left(\dfrac{1}{72}-\dfrac{1}{36}\right)\)
\(=\left(\dfrac{3}{9}-\dfrac{2}{9}\right)-\dfrac{3}{4}+\left(\dfrac{9}{15}+\dfrac{1}{15}\right)+\left(\dfrac{1}{72}-\dfrac{2}{72}\right)\)
\(=\dfrac{1}{9}-\dfrac{3}{4}+\dfrac{10}{15}+\left(-\dfrac{1}{72}\right)\)
\(=\dfrac{1}{9}-\dfrac{3}{4}+\dfrac{2}{3}-\dfrac{1}{72}\)
\(=\dfrac{1}{72}\)
Bài 2.4:
\(A=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)
\(=1-1+\dfrac{1}{72}=\dfrac{1}{72}\)
Bài 2.6:
\(C=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{100}-\dfrac{99}{100}=-\dfrac{98}{100}=-\dfrac{49}{50}\)
