Question

1.So Sánh

a) A=\(\dfrac{11}{2017}+\dfrac{4}{2019}và\) B=\(\dfrac{10}{2017}+\dfrac{10}{2019}\)

b) M=\(\dfrac{1}{5}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{30}+\dfrac{1}{61}+\dfrac{1}{62}và\dfrac{1}{2}\)

c) E=\(\dfrac{4116-14}{10290-35}và\) K=\(\dfrac{2929-101}{2.1919+404}\)

Answer 1

c) E = \(\dfrac{4116-14}{10290-35}\) và K = \(\dfrac{2929-101}{2.1919+404}\)

E = \(\dfrac{4116-14}{10290-35}\)

E = \(\dfrac{14.\left(294-1\right)}{35.\left(294-1\right)}\)

E = \(\dfrac{14}{35}\)

K = \(\dfrac{2929-101}{2.1919+404}\)

K = \(\dfrac{101.\left(29-1\right)}{101.\left(38+4\right)}\)

K = \(\dfrac{29-1}{34+8}\)

K = \(\dfrac{28}{42}\) = \(\dfrac{2}{3}\)

Ta có : E = \(\dfrac{14}{35}\) và K = \(\dfrac{2}{3}\)

\(\dfrac{14}{35}\) = \(\dfrac{42}{105}\)

\(\dfrac{2}{3}\) = \(\dfrac{70}{105}\)

Vậy E < K

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