\(\dfrac{1}{38}+\dfrac{1}{40}+\dfrac{1}{42}+...+\dfrac{1}{50}\) có: \(\left(50-38\right):2+1\)= \(7\) (số hạng)

Ta có: \(\dfrac{1}{38}< \dfrac{1}{7};\dfrac{1}{40}< \dfrac{1}{7};\dfrac{1}{42}< \dfrac{1}{7};...;\dfrac{1}{50}< \dfrac{1}{7}\)

=> \(\dfrac{1}{38}+\dfrac{1}{40}+\dfrac{1}{42}+...+\dfrac{1}{50}< \dfrac{1}{7}+\dfrac{1}{7}+...+\dfrac{1}{7}\)( 7 số hạng)

=> \(\dfrac{1}{38}+\dfrac{1}{40}+\dfrac{1}{42}+...+\dfrac{1}{50}< \dfrac{7}{7}=1\)

Vậy: \(\dfrac{1}{38}+\dfrac{1}{40}+\dfrac{1}{42}+...+\dfrac{1}{50}< 1\)

A= (-1-3-5-7-...-2017)

A= 1+3+5+7+...+2017

A có: (2017-1):2+1=1009 (số hạng)

Tổng A = \(\dfrac{\left(2017+1\right).1009}{2}=1018081\)

A=1018081

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