Vẽ tia \[Ox\], trên tia \[Ox\] lấy hai điểm \[A\] và \[B\] sao cho \[OA = 5\] cm, \[OB = 10\] cm.

a) Chứng minh \[A\] nằm giữa \[O\] và \[B\]. Tính \[AB\].

b) Điểm \[A\] có là trung điểm của \[OB\] không?

c) Vẽ tia \[Ox'\] là tia đối của tia \[Ox\]. Trên tia \[Ox'\] lấy điểm \[C\] sao cho \[OC = 4\] cm. Tính \[BC\].

a) \(\frac{5}{{17}} - \frac{{25}}{{31}} + \frac{{12}}{{17}} + \frac{{ - 6}}{{31}}\)= \(\left( {\frac{5}{{17}} + \frac{{12}}{{17}}} \right) + \left( {\frac{{ - 6}}{{31}} - \frac{{25}}{{31}}} \right)\)

= \(\left( {\frac{{5 + 12}}{{17}}} \right) + \left( {\frac{{ - 6 - 25}}{{31}}} \right)\)= \(\frac{{17}}{{17}} + \frac{{ - 31}}{{31}} = 1 + ( - 1) = 0\).

b) \(\frac{{17}}{8}:\left( {\frac{{23}}{8} + \frac{{ - 9}}{2}} \right)\)=  \(\frac{{17}}{8}:\left( {\frac{{23}}{8} + \frac{{ - 36}}{8}} \right)\)= \(\frac{{17}}{8} & :\frac{{23 + \left( { - 36} \right)}}{8}\)

= \(\frac{{17}}{8} & :\frac{{ - 13}}{8}\)= \(\frac{{17}}{8} & .\frac{8}{{ - 13}} = \frac{{17}}{{ - 13}} = \frac{{ - 17}}{{13}}\).

c) \(\frac{3}{4}.\frac{{11}}{{27}} + \frac{{16}}{{27}}.\frac{3}{4} + \frac{1}{2}\)\[ = \frac{3}{4}.\left( {\frac{{11}}{{27}} + \frac{{16}}{{27}}} \right) + \frac{1}{2}\]

\[ = \frac{3}{4}.\frac{{27}}{{27}} + \frac{1}{2}\]\[ = \frac{3}{4} + \frac{1}{2} = \frac{3}{4} + \frac{2}{4} = \frac{5}{4}\].

d) \(\frac{{27}}{{23}} - \frac{{ - 5}}{{21}} - \frac{4}{{23}} + \frac{{16}}{{21}} + \frac{1}{2}\)\( = \left( {\frac{{27}}{{23}} - \frac{4}{{23}}} \right) + \left( {\frac{{16}}{{21}} - \frac{{ - 5}}{{21}}} \right) + \frac{1}{2}\)

\( = \left( {\frac{{27 - 4}}{{23}}} \right) + \left( {\frac{{16 + 5}}{{21}}} \right) + \frac{1}{2}\)\( = \frac{{23}}{{23}} + \frac{{21}}{{21}} + \frac{1}{2} = 1 + 1 + \frac{1}{2}\)

\( = 2 + \frac{1}{2} = \frac{4}{2} + \frac{1}{2} = \frac{5}{2}\).

Ta có \(A = \frac{4}{{1.2}} + \frac{4}{{2.3}} + \frac{4}{{3.4}} + ... + \frac{4}{{2014.2015}}\)

\[ = 4\,\,.\,\,\left( {\frac{1}{{1\,.\,2}} + \frac{1}{{2\,.\,3}} + \frac{1}{{3\,.\,4}} + ... + \frac{1}{{2014\,.\,2015}}} \right)\]

\( = 4.\left( {1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{2014}} - \frac{1}{{2015}}} \right)\)

\( = 4\,\,.\,\,\left( {1 - \frac{1}{{2015}}} \right)\)\( = 4.\frac{{2014}}{{2015}}\)\( = \frac{{8056}}{{2015}}\).

Vậy \(A = \frac{{8056}}{{2015}}\).