\(A=\frac{3}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{101}\right)\)

\(A=\frac{3}{5}.\left(1-\frac{1}{101}\right)\)

\(A=\frac{3}{5}.\frac{100}{101}\)

\(A=\frac{60}{101}\)

A = 3 - 3/6 + 3/6 - 3/11 + ... + 3/96 - 3/101

A= 3 - 3/101 

A= 300/101

A=\(3.\left(\frac{1}{1.6}\right)+\left(\frac{1}{6.11}\right)+...+\left(\frac{1}{96.101}\right)\)

5A=\(3.\left(\frac{5}{1.6}\right)+\left(\frac{5}{6.11}\right)+...+\left(\frac{5}{96.101}\right)\)

5A=3. \(\left(1-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{11}\right)+...+\left(\frac{1}{96}-\frac{1}{101}\right)\)

5A=3.(1-\(\frac{1}{101}\))

5A=3.\(\frac{100}{101}\)

5A=\(\frac{300}{101}\) suy ra A= \(\frac{300}{101}:5\)=\(\frac{60}{101}\)

`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` 

\(A=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{26.31}\)

\(=5.\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+...+\dfrac{5}{26.31}\right)\)

\(=5.\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)

\(=5.\left(1-\dfrac{1}{31}\right)=5.\dfrac{30}{31}=\dfrac{150}{31}\)

Ta có: \(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{96.101}\) \(=1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\) \(=1-\dfrac{1}{101}\) \(\dfrac{100}{101}\)

\(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+.....+\dfrac{5}{96.101}\)

\(=1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+......+\dfrac{1}{96}-\dfrac{1}{101}\)

\(=1-\dfrac{1}{101}\)

\(=\dfrac{101}{101}-\dfrac{1}{101}\)

\(=\dfrac{101-1}{101}\)

\(=\dfrac{100}{101}\)

`#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.

\(Q=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+......+\dfrac{1}{98.99.100}\)

\(2Q=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+........+\dfrac{2}{98.99.100}\)

\(2Q=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+.....+\dfrac{1}{98.99}-\dfrac{1}{99.100}\)

(do \(\dfrac{2}{a.\left(a+1\right).\left(a+2\right)}=\dfrac{1}{a.\left(a+1\right)}-\dfrac{1}{\left(a+1\right).\left(a+2\right)}\))

\(2Q=\dfrac{1}{1.2}-\dfrac{1}{99.100}\)

\(2Q=\dfrac{1}{2}-\dfrac{1}{9900}=\dfrac{4949}{9900}\)

\(Q=\dfrac{4949}{9900}:2=\dfrac{4949}{19800}\)

Chúc bạn học tốt!!!

\(A=-\dfrac{1}{3}+\dfrac{1}{3^2}-...-\dfrac{1}{3^{99}}+\dfrac{1}{3^{100}}\)

\(=\dfrac{1}{3}\left(-1+\dfrac{1}{3}\right)+\dfrac{1}{3^3}\left(-1+\dfrac{1}{3}\right)+...+\dfrac{1}{3^{99}}\left(-1+\dfrac{1}{3}\right)\)

\(=\dfrac{-2}{3}\left(\dfrac{1}{3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)

Ta có:

\(B=\dfrac{1}{3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)

\(9B=3+\dfrac{1}{3}+...+\dfrac{1}{3^{97}}\)

\(9B-B=3-\dfrac{1}{3^{99}}\)

\(B=\dfrac{3-\dfrac{1}{3^{99}}}{8}\)

\(A=-\dfrac{2}{3}B=\dfrac{-2}{3}.\dfrac{3-\dfrac{1}{99}}{8}=\dfrac{\dfrac{1}{3^{100}}-1}{4}\)

\(S=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+\dfrac{5^2}{11.16}+...+\dfrac{5^2}{96.101}\\ S=\dfrac{25}{1.6}+\dfrac{25}{6.11}+\dfrac{25}{11.16}+...+\dfrac{25}{96.101}\\ S=5.\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{96.101}\right)\\ S=5.\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\right)\\ S=5.\left(1-\dfrac{1}{101}\right)\\ S=5.\dfrac{100}{101}\\ S=\dfrac{500}{101}\)

S=500/101

Huỳnh Huyền Linh làm đúng rùi!

\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)

\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)