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

S=500/101

Huỳnh Huyền Linh làm đúng rùi!

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

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

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

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

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

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

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

Giải:

a) S=5²/1.6+5²/6.11+5²/11.16+5²/16.21+5²/21.26

    S=5.(5.1/6+5/6.11+5/11.16+5/16.21+5/21.26)

    S=5.(1/1-1/6+1/6-1/11+1/11-1/16+1/16-1/21+1/21-1/26)

    S=5.(1/1-1/26)

    S=5.25/26

    S=125/26

b) (1-1/2).(1-1/3).(1-1/4).(1-1/5).....(1-1/19).(1-1/20)

=1/2.2/3.3/4.4/5.....18/19.19/20

=1.2.3.4.....18.19/2.3.4.5.....19.20

=1/20

Chúc bạn học tốt!

Cảm ơn bn

a, bạn tự làm 

b, \(B=\dfrac{5^2}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{101}-\dfrac{1}{106}\right)\)

\(=5\left(1-\dfrac{1}{106}\right)=\dfrac{5.105}{106}=\dfrac{525}{106}\)

c, đk : \(x\ne\dfrac{2}{3}\)

Ta có : \(\left|x-1\right|=2\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)(tm)

Với x = 3 suy ra \(C=\dfrac{2.9+9-1}{3.3-2}=\dfrac{26}{7}\)

Với x = -1 suy ra \(C=\dfrac{2-3-1}{-3-2}=\dfrac{-2}{-5}=\dfrac{2}{5}\)

`#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`

Xin lũi nha nãy làm từ lúc mới đăng á mà lo coi phim :v 

`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` 

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

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

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

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

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

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

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

\(A=5\left(1-\dfrac{1}{31}\right)\)

\(A=5-\dfrac{1}{155}\)

\(A< 5\rightarrowđpcm\)

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

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

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

\(\Leftrightarrow A=5\left(1-\dfrac{1}{31}\right)\)

\(\Leftrightarrow A=5.\dfrac{30}{31}=\dfrac{150}{31}\)

Lời giải:
\(5A=\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+....+\frac{501-496}{496.501}\)

\(=\frac{6}{1.6}-\frac{1}{1.6}+\frac{11}{6.11}-\frac{6}{6.11}+\frac{16}{11.16}-\frac{11}{11.16}+...+\frac{501}{496.501}-\frac{496}{496.501}\)

\(=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+....+\frac{1}{496}-\frac{1}{501}=1-\frac{1}{501}=\frac{500}{501}\)

$\Rightarrow A=\frac{100}{501}$

\(A=\dfrac{1}{5}\left(\dfrac{1}{1.6}+...+\dfrac{1}{496.501}\right)\)

\(A=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\cdot\cdot\cdot+\dfrac{1}{495}-\dfrac{1}{501}\right)\)

\(A=\dfrac{1}{5}\left(1-\dfrac{1}{501}\right)\)

\(A=\dfrac{1}{5}\cdot\dfrac{500}{501}=\dfrac{100}{501}\)

 \(\dfrac{5x}{1.6}+\dfrac{5x}{6.11}+\dfrac{5x}{11.16}+\dfrac{5x}{16.21}+\dfrac{5x}{21.26}+\dfrac{5x}{26.31}=1\)

\(=x\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+\dfrac{5}{16.21}+\dfrac{5}{21.26}+\dfrac{5}{26.31}\right)=1\)

\(=x\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{31}\right)=1\)

\(=x\left(1-\dfrac{1}{31}\right)=1\)

\(\Rightarrow x=1:\left(1-\dfrac{1}{31}\right)=\dfrac{31}{30}\)

=1 NHA

lỗi

mik sửa r nhé