Đáp án:
c1 Ta có :
$\frac{\frac{3}{11}+\frac{3}{29}-\frac{3}{170}+\frac{3}{2015}}{\frac{7}{11}+\frac{7}{29}-\frac{7}{170}+\frac{7}{2015}}$
= $\frac{3.(\frac{1}{11}+\frac{1}{29}-\frac{1}{170}+\frac{1}{2015})}{7.(\frac{1}{11}+\frac{1}{29}-\frac{1}{170}+\frac{1}{2015})}$
= $\frac{3}{7}$
c, (1-$\frac{1}{2}$ )x (1-$\frac{1}{3}$ )-(1-$\frac{1}{4}$ )....(1-$\frac{1}{2003}$ )
= $\frac{1}{2}$ x $\frac{2}{3}$ x $\frac{3}{4}$ x .... x $\frac{2002}{2003}$
= $\frac{1x2x3x....x2002}{2x3x4x...x2003}$
= $\frac{1}{2003}$
Giải thích các bước giải:
