Violympic toán 6

HH

1. tính hợp lí

a. A=\(\dfrac{1}{3}\)+\(\dfrac{-3}{4}\)+\(\dfrac{3}{5}\)+\(\dfrac{1}{57}\)+\(\dfrac{-1}{36}\)+\(\dfrac{1}{15}\)+\(\dfrac{-2}{9}\)

b. B=\(\dfrac{1}{2}\)+\(\dfrac{-1}{5}\)+\(\dfrac{-5}{7}\)+\(\dfrac{1}{6}\)+\(\dfrac{-3}{35}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{41}\)

c. C=\(\dfrac{-1}{2}\)+\(\dfrac{3}{5}\)+\(\dfrac{-1}{9}\)+\(\dfrac{1}{107}\)+\(\dfrac{-7}{18}\)+\(\dfrac{4}{35}\)+\(\dfrac{2}{7}\)

NT
25 tháng 6 2022 lúc 13:14

a: \(A=\left(\dfrac{-3}{4}+\dfrac{-2}{9}-\dfrac{1}{36}\right)+\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{3}{5}\right)+\dfrac{1}{57}\)

\(=\dfrac{-27-8-1}{36}+\dfrac{5+1+9}{15}+\dfrac{1}{57}\)

=1/57

b: \(B=\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{3}\right)+\left(\dfrac{-1}{5}-\dfrac{5}{7}-\dfrac{3}{35}\right)+\dfrac{1}{41}\)

\(=\dfrac{3+1+2}{6}+\dfrac{-7-25-3}{35}+\dfrac{1}{41}\)

=1/41

c: \(C=\left(\dfrac{-1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{2}{7}+\dfrac{4}{35}\right)+\dfrac{1}{107}\)

=1-1+1/107

=1/107

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