Cho A = $\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+\mathrm{\dots }+\frac{4}{2014.2015}$

Tính A.

Lời giải:

\(\frac{1}{2} + \frac{1}{3} = \frac{3}{{2.3}} + \frac{2}{{3.2}} = \frac{3}{6} + \frac{2}{6} = \frac{{3 + 2}}{6} = \frac{5}{6}\)

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

= \(\frac{{17}}{8}:\left( {\frac{{23}}{8} + \frac{{\left( { - 9} \right).4}}{{2.4}}} \right)\)

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

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

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

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

c) \(\frac{3}{4}.\frac{{11}}{{27}} + \frac{{16}}{{27}}.\frac{3}{4} + \frac{1}{2}\)

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

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

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

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

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

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

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

\( = \frac{{23}}{{23}} + \frac{{21}}{{21}} + \frac{1}{2} = 1 + 1 + \frac{1}{2}\)

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