Tìm \(x \in \mathbb{Z},\) biết:

e) \(\frac{{ - 3}}{7} = \frac{{ - 9}}{{2x - 5}}.\)   

a) \(\frac{x}{{ - 5}} = \frac{6}{{ - 10}}\)

Ta có \(\frac{x}{{ - 5}} = \frac{6}{{ - 10}} = \frac{{6:2}}{{\left( { - 10} \right):2}} = \frac{3}{{ - 5}}\)

h) \[\frac{1}{3} + \frac{1}{{15}} + \frac{1}{{35}} + \frac{1}{{63}} + \frac{1}{{99}} + \frac{1}{{143}} + \frac{1}{{195}}\]

\( = \frac{1}{{1 \cdot 3}} + \frac{1}{{3 \cdot 5}} + \frac{1}{{5 \cdot 7}} + \frac{1}{{7 \cdot 9}} + \frac{1}{{9 \cdot 11}} + \frac{1}{{11 \cdot 13}} + \frac{1}{{13 \cdot 15}}\)

\( = \frac{1}{2} \cdot \left( {\frac{2}{{1 \cdot 3}} + \frac{2}{{3 \cdot 5}} + \frac{2}{{5 \cdot 7}} + \frac{2}{{7 \cdot 9}} + \frac{2}{{9 \cdot 11}} + \frac{2}{{11 \cdot 13}} + \frac{2}{{13 \cdot 15}}} \right)\)

\[ = \frac{1}{2} \cdot \left( {\frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + \frac{1}{9} - \frac{1}{{11}} + \frac{1}{{11}} - \frac{1}{{13}} + \frac{1}{{13}} - \frac{1}{{15}}} \right)\]

\[ = \frac{1}{2} \cdot \left( {\frac{1}{1} - \frac{1}{{15}}} \right) = \frac{1}{2} \cdot \left( {\frac{{15}}{{15}} - \frac{1}{{15}}} \right)\]

\[ = \frac{1}{2} \cdot \frac{{14}}{{15}} = \frac{7}{{15}}.\]