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\(T=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{5}\right)\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{8}\right)\left(1-\dfrac{1}{10}\right)\)\(\Rightarrow T=\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}.\dfrac{8}{9}.\dfrac{10}{11}.\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.\dfrac{7}{8}.\dfrac{9}{10}\)
\(\Rightarrow=\dfrac{1}{11}\)
\(\Rightarrow\) Số nghịch đảo của T là \(11\)
`1//([-1]/2)^2 . |+8|-(-1/2)^3:|-1/16|=1/4 .8+1/8 .16=2+2=4`
`2//|-0,25|-(-3/2)^2:1/4+3/4 .2017^0=0,25-2,25.4+0,75.1=0,25-9+0,75=-8,75+0,75-8`
`3//|2/3-5/6|.(3,6:2 2/5)^3=|-1/6|.(3/2)^3=1/6 . 27/8=9/16`
`4//|(-0,5)^2+7/2|.10-(29/30-7/15):(-2017/2018)^0=|1/4+7/2|.10-1/2:1=|15/4|.10-1/2=15/4 .10-1/2=75/2-1/2=37`
`5// 8/3+(3-1/2)^2-|[-7]/3|=8/3+(5/2)^2-7/3=8/3+25/4-7/3=107/12-7/3=79/12`
a) \(\left(\dfrac{2}{3}-\dfrac{1}{2}-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}-\dfrac{1}{7}\right)\)
\(=-\dfrac{1}{6}\cdot\dfrac{17}{28}\)
\(=-\dfrac{17}{168}\)
b) \(\left(\dfrac{15}{21}\div\dfrac{5}{7}\right)\div\left(\dfrac{6}{5}\div2\right)\)
\(=1\div\dfrac{3}{5}\)
\(=\dfrac{5}{3}\)
a: \(\left(\dfrac{4}{9}+\dfrac{1}{3}\right)^2=\dfrac{49}{81}\)
b: \(\left(\dfrac{1}{2}-\dfrac{3}{5}\right)^3=-\dfrac{1}{1000}\)
c: \(\left(-\dfrac{10}{3}\right)^5\cdot\left(-\dfrac{6}{4}\right)^4=-\dfrac{6250}{3}\)
d: \(\left(\dfrac{3}{4}\right)^3:\left(\dfrac{3}{4}\right)^2:\left(-\dfrac{3}{2}\right)^3=-\dfrac{2}{9}\)
A = (\(\dfrac{5}{6}\) - \(\dfrac{4}{5}\)) . 1\(\dfrac{1}{5}\) + \(\dfrac{3}{16}\) : (\(\dfrac{-1}{2}\))3
A = \(\dfrac{1}{30}\) . \(\dfrac{6}{5}\) + \(\dfrac{3}{16}\) : \(\dfrac{-1}{8}\)
A = \(\dfrac{1}{25}\) + \(\dfrac{3}{16}\) . \(\dfrac{-8}{1}\)
A = \(\dfrac{1}{25}\) + \(\dfrac{-3}{2}\)
A = \(\dfrac{-73}{50}\)
B = \(\dfrac{4}{17}\) . (7\(\dfrac{3}{4}\) - 6\(\dfrac{1}{3}\)) + (5\(\dfrac{3}{4}\) - 6.95) : (-1\(\dfrac{3}{5}\))
B = \(\dfrac{4}{17}\) . \(\dfrac{17}{12}\) + (\(\dfrac{23}{4}\) - \(\dfrac{139}{20}\)) : \(\dfrac{-8}{5}\)
B = \(\dfrac{1}{3}\) + \(\dfrac{-6}{5}\) . \(\dfrac{-5}{8}\)
B = \(\dfrac{13}{12}\)
Giải:
\(\dfrac{\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{2}{7}\right)+\left(1-\dfrac{3}{8}\right)+...+\left(1-\dfrac{88}{93}\right)}{\dfrac{-1}{12}-\dfrac{1}{14}-\dfrac{1}{16}-...-\dfrac{1}{186}}\)
Gọi dãy là A,phần tử là B. Ta có:
B=\(\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{2}{7}\right)+\left(1-\dfrac{3}{8}\right)+...+\left(1-\dfrac{88}{93}\right)\)
B=\(\dfrac{5}{6}+\dfrac{5}{7}+\dfrac{5}{8}+...+\dfrac{5}{93}\)
B=5.\(\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+...+\dfrac{1}{93}\right)\)
B=5.\(\left[\dfrac{2}{2}.\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+...+\dfrac{1}{93}\right)\right]\)
B=5.\(\left[2.\left(\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+...+\dfrac{1}{186}\right)\right]\)
B=10.\(\left(\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+...+\dfrac{1}{186}\right)\)
⇒A=\(\dfrac{10.\left(\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+...+\dfrac{1}{186}\right)}{\dfrac{-1}{12}+\dfrac{-1}{14}+\dfrac{-1}{16}+...+\dfrac{-1}{186}}\)
⇒A=-10
Chúc bạn học tốt!
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right).......\left(1-\dfrac{1}{10}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right).........\left(\dfrac{10}{10}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}......\dfrac{9}{10}\)
\(=\dfrac{1}{10}\)