Tính: \(E = \left( {\dfrac{{ - 3}}{{31}} + \dfrac{{ - 6}}{{17}} + \dfrac{1}{{25}}} \right) + \left( {\dfrac{{ - 28}}{{31}} + \dfrac{{ - 11}}{{17}} + \dfrac{{ - 1}}{5}} \right)\)

Câu hỏi :

Tính: \(E = \left( {\dfrac{{ - 3}}{{31}} + \dfrac{{ - 6}}{{17}} + \dfrac{1}{{25}}} \right) + \left( {\dfrac{{ - 28}}{{31}} + \dfrac{{ - 11}}{{17}} + \dfrac{{ - 1}}{5}} \right)\)

A. E = 1

B.  \(E = \dfrac{{ - 54}}{{25}} \)

C.  \(E = \dfrac{{ 54}}{{25}} \)

D. E=0

* Đáp án

B

* Hướng dẫn giải

Ta có: 

\(\begin{array}{l}E = \left( {\dfrac{{ - 3}}{{31}} + \dfrac{{ - 6}}{{17}} + \dfrac{1}{{25}}} \right) + \left( {\dfrac{{ - 28}}{{31}} + \dfrac{{ - 11}}{{17}} + \dfrac{{ - 1}}{5}} \right)\\E = \dfrac{{ - 3}}{{31}} + \dfrac{{ - 6}}{{17}} + \dfrac{1}{{25}} + \dfrac{{ - 28}}{{31}} + \dfrac{{ - 11}}{{17}} + \dfrac{{ - 1}}{5}\\E = \left( {\dfrac{{ - 3}}{{31}} + \dfrac{{ - 28}}{{31}}} \right) + \left( {\dfrac{{ - 6}}{{17}} + \dfrac{{ - 11}}{{17}}} \right) + \left( {\dfrac{1}{{25}} + \dfrac{{ - 1}}{5}} \right)\\E = - 1 - 1 - \dfrac{4}{{25}}\\E = \dfrac{{ - 54}}{{25}}\end{array}\)

 

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