Thực hiện phép tính sau: \(\frac{{1000}}{{1009}} \cdot \frac{{ - 2018}}{{2019}} + \frac{{19}}{{2018}} \cdot \frac{{ - 2018}}{{2019}} + \frac{1}{{202

Câu hỏi :

Thực hiện phép tính: \(\frac{{1000}}{{1009}} \cdot \frac{{ - 2018}}{{2019}} + \frac{{19}}{{2018}} \cdot \frac{{ - 2018}}{{2019}} + \frac{1}{{2020}}\)  

A. \(\frac{{ - 2019}}{{2020}}\)

B. \(\frac{{ 2019}}{{2020}}\)

C. \(\frac{{ - 2020}}{{2019}}\)

D. \(\frac{{ 2020}}{{2019}}\)

* Đáp án

A

* Hướng dẫn giải

\(\begin{array}{l}\,\,\,\,\frac{{1000}}{{1009}} \cdot \frac{{ - 2018}}{{2019}} + \frac{{19}}{{2018}} \cdot \frac{{ - 2018}}{{2019}} + \frac{1}{{2020}}\\ = \left( {\frac{{1000}}{{1009}} \cdot \frac{{ - 2018}}{{2019}} + \frac{{19}}{{2018}} \cdot \frac{{ - 2018}}{{2019}}} \right) + \frac{1}{{2020}}\\ = \frac{{ - 2018}}{{2019}} \cdot \left( {\frac{{1000}}{{1009}} + \frac{{19}}{{2018}}} \right) + \frac{1}{{2020}}\\ = \frac{{ - 2018}}{{2019}} \cdot \left( {\frac{{2000}}{{2018}} + \frac{{19}}{{2018}}} \right) + \frac{1}{{2020}}\\ = \frac{{ - 2018}}{{2019}} \cdot \frac{{2019}}{{2018}} + \frac{1}{{2020}}\\ =  - 1 + \frac{1}{{2020}}\\ = \frac{{ - 2020}}{{2020}} + \frac{1}{{2020}}\\ = \frac{{ - 2019}}{{2020}}\end{array}\)

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