\(\dfrac{1}{1.6}\)+\(\dfrac{1}{6.11}\)+\(\dfrac{1}{11.16}\)+...+\(\dfrac{1}{𝒙.(𝒙+5.\left(\right)}\)=\(\dfrac{26}{131}\).
Ai lm đúng mik kb cho nnha!乁[ᓀ˵▾˵ᓂ]ㄏ
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1 tháng 3 2022
\(VT=\dfrac{1}{5}\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{\left(5n+1\right)\left(5n+6\right)}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{5n+6}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{5n+6-1}{5n+6}\)
\(=\dfrac{n+1}{5n+6}=VP\)
11 tháng 2 2022
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 \(C=\dfrac{2-3-1}{-3-2}=\dfrac{-2}{-5}=\dfrac{2}{5}\)
TK
2
AH
Akai Haruma
Giáo viên
20 tháng 3 2022
Lời giải:
\(5A=\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+....+\frac{501-496}{496.501}\)
\(=\frac{6}{1.6}-\frac{1}{1.6}+\frac{11}{6.11}-\frac{6}{6.11}+\frac{16}{11.16}-\frac{11}{11.16}+...+\frac{501}{496.501}-\frac{496}{496.501}\)
\(=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+....+\frac{1}{496}-\frac{1}{501}=1-\frac{1}{501}=\frac{500}{501}\)
$\Rightarrow A=\frac{100}{501}$
20 tháng 3 2022
\(A=\dfrac{1}{5}\left(\dfrac{1}{1.6}+...+\dfrac{1}{496.501}\right)\)
\(A=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\cdot\cdot\cdot+\dfrac{1}{495}-\dfrac{1}{501}\right)\)
\(A=\dfrac{1}{5}\left(1-\dfrac{1}{501}\right)\)
\(A=\dfrac{1}{5}\cdot\dfrac{500}{501}=\dfrac{100}{501}\)
2 tháng 5 2021
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!
AH
Akai Haruma
Giáo viên
15 tháng 4 2023
a.
$A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{1000-999}{999.1000}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}$
$=1-\frac{1}{1000}=\frac{999}{1000}$
AH
Akai Haruma
Giáo viên
15 tháng 4 2023
b.
$5B=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+....+\frac{5}{495.500}$
$=\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+....+\frac{500-495}{495.500}$
$=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{495}-\frac{1}{500}$
$=1-\frac{1}{500}=\frac{499}{500}$
$\Rightarrow B=\frac{499}{500}: 5= \frac{499}{2500}$
C
12 tháng 2 2023
`#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`
1 tháng 4 2017
\(\dfrac{1}{1\cdot6}+\dfrac{1}{6\cdot11}+\dfrac{1}{11\cdot16}+...+\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}=\dfrac{n+1}{5n+6}\)
\(VT=\dfrac{1}{5}\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{\left(5n+1\right)\left(5n+6\right)}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{5n+6}\right)\)\(=\dfrac{1}{5}\cdot\left(\dfrac{5n+6}{5n+6}-\dfrac{1}{5n+6}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{5\left(n+1\right)}{5n+6}=\dfrac{n+1}{5n+6}=VP\)
VV
25 tháng 10 2021
\(1\)/
\(a\)) \(=\left(\dfrac{7}{5}-\dfrac{8}{7}\right)+\dfrac{17}{5}:0,6\)
\(=\dfrac{9}{35}+\dfrac{17}{3}\)
\(=\dfrac{622}{105}\)
\(b\)) \(=\dfrac{11}{6}+\dfrac{-14}{15}\)
\(=\dfrac{9}{10}\)
\(c\)/ \(=\dfrac{7}{4}-\dfrac{2}{3}\)
\(=\dfrac{13}{12}\)
TT
2
19 tháng 11 2017
Ta có: \(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{96.101}\) \(=1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\) \(=1-\dfrac{1}{101}\) \(\dfrac{100}{101}\)
NN
19 tháng 11 2017
\(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+.....+\dfrac{5}{96.101}\)
\(=1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+......+\dfrac{1}{96}-\dfrac{1}{101}\)
\(=1-\dfrac{1}{101}\)
\(=\dfrac{101}{101}-\dfrac{1}{101}\)
\(=\dfrac{101-1}{101}\)
\(=\dfrac{100}{101}\)
\(\dfrac{1}{1\cdot6}+\dfrac{1}{6\cdot11}+...+\dfrac{1}{x\left(x+5\right)}=\dfrac{26}{131}\)
=>\(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{x\left(x+5\right)}=\dfrac{130}{131}\)
=>\(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+5}=\dfrac{130}{131}\)
=>\(1-\dfrac{1}{x+5}=\dfrac{130}{131}\)
=>\(\dfrac{1}{x+5}=1-\dfrac{130}{131}=\dfrac{1}{131}\)
=>x+5=131
=>x=126(nhận)
Ê mà nhiều lần mik thấy bn nhiều lắm đó. Thịnh