`#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 = 2(5/1.6 + 5/6.11 +.......+ 5/101.106)

S = 2( 1 - 1/6 + 1/6 - 1/11 +.....+ 1/101 - 1/106)

S = 2( 1 - 1/106)

S = 2 . 105/106

S = 105/53

k mk đi,mk mới bị trừ điểm!

1/2.S=5/(1.6)+5/(6.11)+...+5/(101.106)

1/2.S=1/1-1/6+1/6-1/11+...+1/101-1/106

1/2.S=1/1-1/106

1/2.S=105/106

S=105/53

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 \(C=\dfrac{2-3-1}{-3-2}=\dfrac{-2}{-5}=\dfrac{2}{5}\)

\(S=2\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{101.106}\right)\)

\(S=2\left(1-\frac{1}{106}\right)\)

\(S=\frac{210}{106}=\frac{105}{53}\)

\(S=2.\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{101.106}\right)\)

\(S=2.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-...-\frac{1}{101}+\frac{1}{101}-\frac{1}{106}\right)\)

\(S=2.\left[1+\left(\frac{-1}{6}+\frac{1}{6}\right)+\left(\frac{-1}{11}\frac{1}{11}\right)+...+\left(\frac{-1}{101}+\frac{1}{101}\right)-\frac{1}{106}\right]\)

\(S=2.\left[1+0+0+...+0-\frac{1}{106}\right]\)

\(S=2.\left[1-\frac{1}{106}\right]\)

\(S=2.\frac{105}{106}\)

\(S=\frac{10}{1.6}+\frac{10}{6.11}+...+\frac{10}{101.106}\)

=\(2\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{101.106}\right)\)

\(=2\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{101}+\frac{1}{106}\right)\)

\(=2\left(1-\frac{1}{106}\right)=2.\frac{105}{106}=\frac{105}{53}\)

A=1/15-1/16+1/16-1/17+...+1/2016-1/2017

A=1/15-1/2017

A=2002/30255

C=1/3[3/5.8+3/8.11+...+3/101.104]

C=1/3[1/5-1/8+1/8-1/11+...+1/101-1/104]

C=1/3[1/5-1/104]

C=1/3.99/520

C=33/520

D=2/3[3/1.4+3/4.7+.....+3/101.104]

D=2/3[1-1/4+1/4-1/7+.....+1/101-1/104]

D=2/3[1-1/104]

D=2/3.103/104

D=103/156

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!

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

\(\dfrac{31}{30}\)bạn nhé

Chúc Bạn Học Tốt.

lỗi