Bài 2:
a: \(A=\dfrac{3}{4}-\dfrac{1}{4}-\dfrac{5}{6}-\dfrac{7}{3}+\dfrac{9}{2}=5-\dfrac{5}{6}-\dfrac{14}{6}=5-\dfrac{19}{6}=\dfrac{11}{6}\)
b: \(B=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)
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