Đáp án: Tính a) 2/3 ×5 + 7/5 ×12 +9/4 ×39 b)(1-1/1 ×2) + (1-1/2 ×3)++(1-1/2013 ×2014) c)-1/2003 ×2002 -1/2002 ×2001--1/2 ×1 - [(l) (2/(35)) + (7/(512)) + (9/(439)) = (1/5)( ((2/3) + (7/(12))) ) + (9/(439)) = (1/5)(5/4) + (9/(439)) = (1/4) + (9/(439)) = (1/4)( (1 + (9/(39))) ) = (1/4)((16)/(13)) = (4/(13)) ] [(l) b) 1 - (1/(12)) + 1 - (1/(23)) + + 1 - (1/(20132014)) = 1 - ( (1 - (1/2)) ) + 1 - ( ((1/2) - (1/3)) ) + + 1 -...

[(l) (2/(35)) + (7/(512)) + (9/(439)) = (1/5)( ((2/3) + (7/(12))) ) + (9/(439)) = (1/5)(5/4) + (9/(439)) = (1/4) + (9/(439)) = (1/4)( (1 + (9/(39))) ) = (1/4)((16)/(13)) = (4/(13)) ] [(l) b) 1 - (1/(12)) + 1 - (1/(23)) + + 1 - (1/(20132014)) = 1 - ( (1 - (1/2)) ) + 1 - ( ((1/2) - (1/3)) ) + + 1 -...

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2 năm trước

Tính a) 2/3 ×5 + 7/5 ×12 +9/4 ×39 b)(1-1/1 ×2) + (1-1/2 ×3)+....+(1-1/2013 ×2014) c)-1/2003 ×2002 -1/2002 ×2001-...-1/2 ×1

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thutrangdoan289

2 năm trước

\[\begin{array}{l} \frac{2}{{3.5}} + \frac{7}{{5.12}} + \frac{9}{{4.39}} = \frac{1}{5}\left( {\frac{2}{3} + \frac{7}{{12}}} \right) + \frac{9}{{4.39}}\\ = \frac{1}{5}.\frac{5}{4} + \frac{9}{{4.39}} = \frac{1}{4} + \frac{9}{{4.39}} = \frac{1}{4}\left( {1 + \frac{9}{{39}}} \right)\\ = \frac{1}{4}.\frac{{16}}{{13}} = \frac{4}{{13}}. \end{array}\] \[\begin{array}{l} b)\,\,1 - \frac{1}{{1.2}} + 1 - \frac{1}{{2.3}} + ...... + 1 - \frac{1}{{2013.2014}}\\ = 1 - \left( {1 - \frac{1}{2}} \right) + 1 - \left( {\frac{1}{2} - \frac{1}{3}} \right) + ..... + 1 - \left( {\frac{1}{{2013}} - \frac{1}{{2014}}} \right)\\ = 1 - 1 + \frac{1}{2} + 1 - \frac{1}{2} + \frac{1}{3} + ..... + 1 - \frac{1}{{2013}} + \frac{1}{{2014}}\\ = \underbrace {1 + 1 + ... + 1}_{2012\,\,\,chu\,\,so\,\,1} + \frac{1}{{2014}} = 2012 + \frac{1}{{2014}} = \frac{{2012.2014 + 1}}{{2014}}. \end{array}\] Câu c bạn làm tương tự.