Bài tập 11.7 trang 28 SBT Toán 6 Tập 2

Ta có nhận xét:

\(\begin{array}{l}
\frac{1}{{1.2}} - \frac{1}{{2.3}} = \frac{{3 - 1}}{{1.2.3}} = \frac{2}{{1.2.3}}\\
\frac{1}{{2.3}} - \frac{1}{{3.4}} = \frac{{4 - 2}}{{2.3.4}} = \frac{2}{{2.3.4;}}...
\end{array}\)

Suy ra

\(\begin{array}{l}
\frac{1}{{1.2.3}} = \frac{1}{2}\left( {\frac{1}{{1.2}} - \frac{1}{{2.3}}} \right)\\
\frac{1}{{2.3.4}} = \frac{1}{2}.\left( {\frac{1}{{2.3}} - \frac{1}{{3.4}}} \right);...
\end{array}\)

Do đó:

\(\begin{array}{l}
M = \frac{1}{2}.\left( {\frac{1}{{1.2}} - \frac{1}{{2.3}} + \frac{1}{{2.3}} - \frac{1}{{3.4}} + ... + \frac{1}{{10.11}} - \frac{1}{{11.12}}} \right)\\
 = \frac{1}{2}.\left( {\frac{1}{{1.2}} - \frac{1}{{11.12}}} \right)\\
 = \frac{1}{2}.\left( {\frac{1}{2} -  - \frac{1}{{11.12}}} \right)\\
 = \frac{1}{2}.\frac{{65}}{{132}} = \frac{{65}}{{264}}
\end{array}\)

-- Mod Toán 6 HỌC247