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Ta có : A = 1/ 2.5 + 1/ 5.8 + 1/ 8.11 + ... + 1/ (3n-1).(3n+2) .
= 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/ 3n-1 - 1/ 3n+2 .
= 1/2 - 1/ 3n+2 .
= 3n + 2 - 2 / 2 .( 3n+2 ) .
= 3n / 2.(3n+2) .
\(a,\left(\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|\right):10=\left(1-\frac{1}{2}\right)....\left(1-\frac{1}{10}\right)\)
\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\Leftrightarrow\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|=1\)
\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.|x-2|=1\Leftrightarrow|x-2|.\frac{2}{3}=1\Leftrightarrow|x-2|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)
\(\left(\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|\right):10=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{9}\right).\left(1-\frac{1}{10}\right)\)
\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\)
\(\Leftrightarrow\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|=1\)
\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.\left|x-2\right|=1\)
\(\Leftrightarrow\left|x-2\right|.\frac{2}{3}=1\Leftrightarrow\left|x-2\right|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)
Ta có:
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(3n+2\right).\left(3n+5\right)}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{\left(3n+2\right).\left(3n+5\right)}\right)\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n+2}-\frac{1}{3n+5}\right)\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3n+5}\right)\)
\(\Rightarrow\frac{1}{6}-\frac{1}{9n+15}\)
a) \(\frac{-7}{15}.\frac{5}{8}.\frac{15}{-7}.\left(-16\right)\)
\(=\left(\frac{-7}{15}.\frac{15}{-7}\right).\left[\frac{15}{8}.\left(-16\right)\right]\)
\(=1.\left[15.\left(-2\right)\right]=1.\left(-30\right)=-30\)
b) \(\left(-\frac{1}{2}\right).3\frac{1}{5}+\left(-\frac{1}{2}\right).\left(-2\frac{1}{5}\right)\)
\(=\left(-\frac{1}{2}\right).\left(3\frac{1}{5}-2\frac{1}{5}\right)\)
\(=\left(-\frac{1}{2}\right).1=-\frac{1}{2}\)
a) A = \(\frac{15}{7}:\left(\frac{1}{15}-\frac{7}{5}\right)-\frac{15}{7}:\left(\frac{17}{15}+\frac{11}{5}\right)=\frac{15}{7}:\frac{-20}{15}-\frac{15}{7}:\frac{50}{15}\)
A = \(\frac{15}{7}.\frac{15}{-20}-\frac{15}{7}.\frac{15}{50}=\frac{15}{7}.\left(\frac{-15}{20}-\frac{15}{50}\right)=\frac{15}{7}.\frac{-105}{100}=-\frac{9}{4}\)
b) B = \(\frac{1}{\left(-\frac{2}{3}\right)^4}.\left(-4\right)^2-1^{2016}-10\frac{1}{3}=\frac{1}{\frac{16}{81}}.16-1-10\frac{1}{3}=\frac{81}{16}.16-1-10\frac{1}{3}\)
B = \(81-1-10-\frac{1}{3}=70-\frac{1}{3}=\frac{209}{3}\)
