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\(A=\frac{3}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{101}\right)\)
\(A=\frac{3}{5}.\left(1-\frac{1}{101}\right)\)
\(A=\frac{3}{5}.\frac{100}{101}\)
\(A=\frac{60}{101}\)
A = 3 - 3/6 + 3/6 - 3/11 + ... + 3/96 - 3/101
A= 3 - 3/101
A= 300/101
`A = ( 5^2 )/( 1*6)+(5^2)/(6*11)+.....+(5^2)/(26*31)`
`= 5*( 5/( 1*6)+ 5/(6*11)+.....+5/(26*31))`
`= 5*( 1 - 1/6 + 1/6 - 1/11 +....+1/26 - 1/31 )`
`= 5*( 1 - 1/31 )`
`= 5 * 30/31 = 150/31`
\(A=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{26.31}\)
\(=5.\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+...+\dfrac{5}{26.31}\right)\)
\(=5.\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
\(=5.\left(1-\dfrac{1}{31}\right)=5.\dfrac{30}{31}=\dfrac{150}{31}\)
`#lv`
`A=(-1)+(-5)+(-9)+...+(-101)`
`=-(1+5+9+...+101)`
Số số hạng là :
`[101-(-1)]:4+1=26(` số hạng `)`
Tổng là :
`[(-101)+(-1)]xx26:2=-1326`
Vậy `A=-1326`
__
`B=-5/17 . 8/19 + (-12)/17 . 8/19 - 11/19`
`=((-5)/17+(-12)/17).8/19-11/19`
`=-1.8/19-11/19`
`=-8/19-11/19`
`=-8/19+(-11)/19`
`=-19/19`
`=-1`
__
`C=10/1.6 + 10/6.11 + 10/11.16 + ... + 10/2016.2021`
`=2.(1-1/6+1/6-1/11+...+1/2016-1/2021)`
`=2(1-1/2021)`
`=2. (2021/2021-1/2021)`
`=2. 2020/2021`
`=4040/2021`
S = \(\dfrac{1}{1.4}\)+ \(\dfrac{1}{4.7}\)+...+\(\dfrac{1}{2002.2005}\)
S = ( 1 - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-...+\(\dfrac{1}{2002}\)-\(\dfrac{1}{2005}\)) . \(\dfrac{1}{3}\)
S = ( 1 - \(\dfrac{1}{2005}\)) . \(\dfrac{1}{3}\)
S = \(\dfrac{2004}{2005}\). \(\dfrac{1}{3}\)
S = \(\dfrac{2014}{6015}\)
a) \(S=\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{2002.2005}\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2002}-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{3}.\dfrac{2004}{2005}=\dfrac{668}{2005}\)
KL.
b) \(P=\dfrac{3}{1.6}+\dfrac{3}{6.11}+\dfrac{3}{11.16}+...+\dfrac{3}{96.101}\)
\(=\dfrac{3}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{5}\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{5}.\dfrac{100}{101}=\dfrac{60}{101}\)
KL.
c) \(Q=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{9900}\right)\)
\(=\dfrac{1}{2}.\dfrac{1}{19800}=\dfrac{1}{39600}\)
KL.
\(S=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+\dfrac{5^2}{11.16}+...+\dfrac{5^2}{96.101}\\ S=\dfrac{25}{1.6}+\dfrac{25}{6.11}+\dfrac{25}{11.16}+...+\dfrac{25}{96.101}\\ S=5.\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{96.101}\right)\\ S=5.\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\right)\\ S=5.\left(1-\dfrac{1}{101}\right)\\ S=5.\dfrac{100}{101}\\ S=\dfrac{500}{101}\)
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)
\(=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}.\dfrac{12}{7}\)
\(=\dfrac{2}{3}+\dfrac{7.3.2.2}{3.7.3.2.3}\)
\(=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)
TICK CHO MÌNH NHÉ
Giải:
\(\dfrac{2}{3}\) + \(\dfrac{1}{3}\) . (\(-\dfrac{4}{9}\) + \(\dfrac{5}{6}\) ) : \(\dfrac{7}{12}\)
= \(\dfrac{2}{3}\) + \(^{\dfrac{1}{3}}\) . \(\dfrac{7}{18}\) : \(\dfrac{7}{12}\)
= \(\dfrac{2}{3}\) + \(\dfrac{7}{54}\) : \(\dfrac{7}{12}\)
= \(\dfrac{2}{3}\) + \(\dfrac{2}{9}\)
= \(\dfrac{8}{9}\)
P= \(\dfrac{3}{1.6}\)+\(\dfrac{3}{6.11}\)+\(\dfrac{3}{11.17}\)+...+\(\dfrac{3}{96.101}\)
\(\dfrac{5}{3}\).P= \(\dfrac{5}{3}\).(\(\dfrac{3}{1.6}\)+\(\dfrac{3}{6.11}\)+\(\dfrac{3}{11.16}\)+...+\(\dfrac{3}{96.101}\))
\(\dfrac{5}{3}\).P= \(\dfrac{5}{1.6}\)+\(\dfrac{5}{6.11}\)+\(\dfrac{5}{11.16}\)+...+\(\dfrac{5}{96.101}\)
\(\dfrac{5}{3}\).P= \(\dfrac{1}{1}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{16}\)+...+\(\dfrac{1}{96}\)-\(\dfrac{1}{101}\)
\(\dfrac{5}{3}\).P= \(\dfrac{1}{1}\)-\(\dfrac{1}{101}\)= \(\dfrac{101}{101}\)-\(\dfrac{1}{101}\)=\(\dfrac{100}{101}\)
P= \(\dfrac{100}{101}\):\(\dfrac{5}{3}\)= \(\dfrac{100}{101}\).\(\dfrac{3}{5}\)=\(\dfrac{100.3}{101.5}\)=\(\dfrac{20.3}{101.1}\)=\(\dfrac{60}{101}\)
Vậy P= \(\dfrac{60}{101}\)
11.17 chứ không phải 11.16 bạn àk