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