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\(\frac{4}{3.6}+\frac{4}{6.9}+\frac{4}{9.12}+\frac{4}{12.15}\)
\(=\frac{4}{3}.\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+\frac{3}{12.15}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\frac{4}{3}.\left(\frac{5}{15}-\frac{1}{15}\right)\)
\(=\frac{4}{3}.\frac{4}{15}=\frac{16}{45}\)
Dấu . là nhân nha
\(\frac{4}{3.6}+\frac{4}{6.9}+\frac{4}{9.12}+\frac{4}{12.15}\)
\(=\frac{4}{3}.\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+\frac{3}{12.15}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\frac{4}{3}.\frac{4}{15}=\frac{16}{45}\)
\(\frac{4}{3.6}+\frac{4}{6.9}+\frac{4}{9.12}+\frac{4}{12.15}\)
\(=\frac{4}{3}\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+\frac{3}{12.15}\right)\)
\(=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\right)\)
\(=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\frac{4}{3}.\frac{4}{15}=\frac{16}{45}\)
Ta có :
1/2x6 + 1/4x9 + 1/6x12 +...+1/198 x 300
= 1/6x2 + 1/6x6 + 1/6x12 + ....+1/6x9900
= 1/6 x ( 1/2 + 1/6 + 1/ 12 +...+1/9900)
= 1/6 x (1/1x2 + 1/2x3 + 1/3x4+...+1/99x100)
=1/6x (1-1/2 + 1/2-1/3 + 1/3 - 1/4 + ....+1/99-1/100)
=1/6x(1-1/100)
=1/6 x 99/100
= 33/200
k cho mình nha , học tốt
\(\frac{1}{2.6}+\frac{1}{4.9}+\frac{1}{6.12}+...+\frac{1}{36.57}+\frac{1}{38.60}\)
\(=\frac{1}{2.3}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(=\frac{1}{6}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=\frac{1}{6}.\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{6}.\frac{19}{20}=\frac{19}{120}\)
a) \(T=\frac{9^{14}\times25^6\times8^7}{18^{12}\times625^3\times24^3}\)
\(=\frac{\left(3^2\right)^{14}\times25^6\times\left(2^3\right)^7}{\left(2\times3^2\right)^{12}\times\left(25^2\right)^3\times\left(3\times2^3\right)^3}\)
\(=\frac{3^{28}\times25^6\times2^{21}}{2^{12}\times3^{24}\times25^6\times3^3\times2^9}\)
\(=\frac{3^{28}\times25^6\times2^{21}}{\left(2^{12}\times2^9\right)\times\left(3^{24}\times3^3\right)\times25^6}\)
\(=\frac{3^{28}\times25^6\times2^{21}}{2^{21}\times3^{27}\times25^6}=3\)
b) \(A=\frac{5\times4^{15}\times9^9-4\times3^{20}\times8^9}{5\times2^9\times6^{19}-7\times2^{29}\times27^6}\)
\(=\frac{5\times\left(2^2\right)^{15}\times\left(3^2\right)^9-2^2\times3^{20}\times\left(2^3\right)^9}{5\times2^9\times\left(2\times3\right)^{19}-7\times2^{29}\times\left(3^3\right)^6}\)
\(=\frac{5\times2^{30}\times3^{18}-2^2\times3^{20}\times2^{27}}{5\times2^9\times2^{19}\times3^{19}-7\times2^{29}\times3^{18}}\)
\(=\frac{5\times2^{30}\times3^{18}-2^{29}\times3^{20}}{5\times2^{28}\times3^{19}-7\times2^{29}\times3^{18}}\)
\(=\frac{2^{29}\times3^{18}\times\left(5\times2-3^2\right)}{2^{28}\times3^{18}\times\left(5\times3-7\times2\right)}\)
\(=\frac{2\times\left(10-9\right)}{15-14}=\frac{2\times1}{1}=2\)
\(A=\dfrac{1}{5\times7}+\dfrac{1}{7\times9}+\dfrac{1}{9\times11}+...+\dfrac{1}{87\times89}\)
\(A=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{87}-\dfrac{1}{89}\)
\(A=\dfrac{1}{5}-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-\left(\dfrac{1}{9}-\dfrac{1}{9}\right)-...-\left(\dfrac{1}{87}-\dfrac{1}{87}\right)-\dfrac{1}{89}\)
\(A=\dfrac{1}{5}-\dfrac{1}{89}\)
\(A=\dfrac{84}{445}\)
Vậy, `A=84/445.`
A = \(\dfrac{1}{5\times7}\) + \(\dfrac{1}{7\times9}\)+\(\dfrac{1}{9\times11}\)+...+\(\dfrac{1}{87\times89}\)
A = \(\dfrac{1}{2}\) \(\times\)( \(\dfrac{2}{5\times7}\)+\(\dfrac{2}{7\times9}\)+\(\dfrac{2}{9\times11}\)+...+\(\dfrac{2}{87\times89}\))
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{11}\) +...+ \(\dfrac{1}{87}\) - \(\dfrac{1}{89}\))
A = \(\dfrac{1}{2}\) \(\times\) (\(\dfrac{1}{5}\) - \(\dfrac{1}{89}\))
A = \(\dfrac{1}{2}\) \(\times\) \(\dfrac{84}{445}\)
A = \(\dfrac{42}{445}\)
Ta có: \(\frac{1}{4\times6}=\frac{1}{4\times1\times3\times2}=\frac{1}{4\times3\times1\times2}\)
\(\frac{1}{8\times9}=\frac{1}{4\times2\times3\times3}=\frac{1}{4\times3\times2\times3}\)
\(\frac{1}{12\times12}=\frac{1}{4\times3\times3\times4}\)
\(\frac{1}{16\times15}=\frac{1}{4\times4\times3\times5}=\frac{1}{4\times3\times4\times5}\)......
\(\frac{1}{2680\times2013}=\frac{1}{4\times670\times3\times671}\)
Do đó:
\(M=\frac{1}{4\times3}\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{670\times671}\right)\)
\(=\frac{1}{12}\times\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{670}-\frac{1}{671}\right)\)
\(=\frac{1}{12}\times\left(\frac{1}{1}-\frac{1}{671}\right)=\frac{1}{12}\times\frac{670}{671}=\frac{335}{4026}\)
Vậy \(M=\frac{335}{4026}\)
A = \(\dfrac{1}{3\times6}\) + \(\dfrac{1}{6\times9}\) + \(\dfrac{1}{9\times12}\)+...+\(\dfrac{1}{144\times147}\)
A = \(\dfrac{1}{3}\) \(\times\)( \(\dfrac{3}{3\times6}\) + \(\dfrac{3}{6\times9}\)+\(\dfrac{1}{9\times12}\)+...+\(\dfrac{3}{144\times147}\))
A = \(\dfrac{1}{3}\) \(\times\)(\(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{144}-\dfrac{1}{147}\))
A = \(\dfrac{1}{3}\)\(\times\)(\(\dfrac{1}{3}\) - \(\dfrac{1}{147}\))
A = \(\dfrac{1}{3}\) \(\times\)\(\dfrac{16}{49}\)
A = \(\dfrac{16}{147}\)