Bài tập về thực hiện phép cộng nhiều hơn hai phân số

3. a) A = $\frac{124}{125} + \left ( \frac{-18}{15}+\frac{1}{125} \right )+\frac{3}{15}$

     = $\left (\frac{124}{125}+\frac{1}{125}  \right ) + \left ( \frac{-18}{15}++\frac{3}{15} \right )$

     = $\frac{124 + 1}{125} + \frac{-18 + 3}{15}$

     = 1 + (-1)

     = 0

b) B = $\left ( \frac{1}{9}-\frac{5}{17} \right )+\frac{3}{6}+\left ( \frac{-12}{17}+\frac{-1}{2} \right )+\frac{5}{9}$

     = $\left ( \frac{1}{9}+\frac{5}{9} \right ) + \left ( \frac{-12}{17}-\frac{5}{17} \right ) + \left ( \frac{3}{6}+\frac{-1}{2} \right )$

     = $\frac{1+5}{9} + \frac{-12 - 5}{-17} + \frac{1+(-1)}{2}$

     = $\frac{2}{3}-1 +0$

     = $\frac{-1}{3}$

4.  a) Với $(a\in N*)$, ta có:

$\frac{1}{a(a+1)}=\frac{(a+1)+(-a)}{a(a+1)}=\frac{a+1}{a(a+1)}+\frac{-a}{a(a+1)}=\frac{1}{a}+\frac{-1}{a+1}$ (đpcm)

b) C = $\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}$

       = $\frac{1}{1}+\frac{-1}{2}+\frac{1}{2}+\frac{-1}{3}+\frac{1}{3}+\frac{-1}{4}+...+\frac{1}{999}+\frac{-1}{1000}$

       = 1 + $\left ( \frac{-1}{2}+\frac{1}{2} \right ) + \left ( \frac{-1}{3}+\frac{1}{3}+ \right )+ ...+ \left ( \frac{-1}{999}+\frac{1}{999}+ \right ) + \frac{-1}{1000}$

       = $1 + \frac{-1}{1000}$ = $\frac{999}{1000}$