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10 tháng 8 2018
\(\frac{5^2\times6^{11}\times16^2+6^2\times12^6\times15^2}{2\times6^{12}\times10^4-81^2\times960^3}\)
\(=\frac{5^2\times\left(2\times3\right)^{11}\times\left(2^4\right)^2+\left(2\times3\right)^2\times\left(2^2\times3\right)^6\times\left(3\times5\right)^2}{2\times\left(2\times3\right)^{12}\times\left(2\times5\right)^4-\left(3^4\right)^2\times\left(2^6\times3\times5\right)^3}\)
\(=\frac{5^2\times2^{19}\times3^{11}+2^{14}\times3^{10}\times5^3}{2^{17}\times5^4\times3^{12}-3^{11}\times2^{18}\times5^3}\)
\(=\frac{5^2\times3^{10}\times2^{14}\times\left(2^5\times3+5\right)}{2^{17}\times5^3\times3^{11}\times\left(5\times3-2\right)}\)
\(=\frac{2^5\times3+5}{2^3\times5\times3\times12}\)
\(=\frac{32\times3+5}{8\times15\times12}=\frac{96+5}{120\times12}=\frac{101}{1440}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
6 tháng 6 2023
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 = \(\dfrac{16}{147}\)
6 tháng 8 2016
\(\frac{1}{2.6}+\frac{1}{4.9}+\frac{1}{6.12}+...+\frac{1}{36.57}+\frac{1}{38.60}\)
\(=\frac{1}{2.3}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(=\frac{1}{6}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=\frac{1}{6}.\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{6}.\frac{19}{20}=\frac{19}{120}\)
1 tháng 5 2015
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
21 tháng 3 2020
Ta có:
\(\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+.....+\frac{1}{14\times15}=\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+\frac{6-5}{6\times5}+...+\frac{15-14}{14\times15}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{14}-\frac{1}{15}=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
Do đó: \(\frac{4}{15}=\frac{x}{30}\)
\(x=\frac{4}{15}\times30=8\)
21 tháng 3 2020
\(\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{14\times15}=\frac{x}{30}\)
\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{5}+...+\frac{1}{14}-\frac{1}{15}=\frac{x}{30}\)
\(\frac{1}{3}-\frac{1}{15}=\frac{x}{30}\)
\(\frac{4}{15}=\frac{x}{30}\)
\(\Rightarrow4\times30=15\times x\)
\(120=15\times x\)
\(x=120\div15\)
\(x=8.\)
21 tháng 3 2020
\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{14.15}=\frac{x}{30}\)
\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{14}-\frac{1}{15}=\frac{x}{30}\)
\(\frac{1}{3}-\frac{1}{15}=\frac{x}{30}\)
\(\frac{5}{15}-\frac{1}{15}=\frac{x}{30}\)
\(\frac{4}{15}=\frac{x}{30}\)
\(\frac{8}{30}=\frac{x}{30}\)
\(x=8\)
hok tốt!!
24 tháng 5 2019
\(A=\frac{1}{1\times6\times6}+\frac{1}{2\times9\times8}+\frac{1}{3\times12\times10}+...+\frac{1}{98\times297\times200}\)
\(A=\frac{1}{1\times\left(2\times3\right)\times\left(2\times3\right)}+\frac{1}{2\times\left(3\times3\right)\times\left(2\times4\right)}+...+\frac{1}{98\times\left(99\times3\right)\times\left(100\times2\right)}\)
\(A=\frac{1}{6}\times\left(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+...+\frac{1}{98\times99\times100}\right)\)
\(12\times A=\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+...+\frac{2}{98\times99\times100}\)
\(12\times A=\left(\frac{1}{1\times2}-\frac{1}{2\times3}\right)+\left(\frac{1}{2\times3}-\frac{1}{3\times4}\right)+...+\left(\frac{1}{98\times99}-\frac{1}{99\times100}\right)\)
\(12\times A=\frac{1}{1\times2}-\frac{1}{99\times100}=\frac{4949}{9900}\)
\(A=\frac{4949}{118800}\)
\(\frac{4}{3.6}+\frac{4}{6.9}+\frac{4}{9.12}+\frac{4}{12.15}\)
\(=\frac{4}{3}.\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+\frac{3}{12.15}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\frac{4}{3}.\left(\frac{5}{15}-\frac{1}{15}\right)\)
\(=\frac{4}{3}.\frac{4}{15}=\frac{16}{45}\)
Dấu . là nhân nha
\(\frac{4}{3.6}+\frac{4}{6.9}+\frac{4}{9.12}+\frac{4}{12.15}\)
\(=\frac{4}{3}.\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+\frac{3}{12.15}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\frac{4}{3}.\frac{4}{15}=\frac{16}{45}\)