a) 5n+ 5n+1=30
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BX
0
DN
0
9 tháng 8 2019
Ta có : \(A=\frac{1}{1\cdot6}+\frac{1}{6\cdot11}+\frac{1}{11\cdot16}+...+\frac{1}{(5n+1)(5n+6)}\)
\(=\frac{1}{5}\cdot\left[\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{5}{(5n+1)(5n+6)}\right]\)
\(=\frac{1}{5}\cdot\left[1-\frac{1}{5n+6}\right]=\frac{1}{5}\cdot\frac{5n+6-1}{5n+6}=\frac{1}{5}\cdot\frac{5(n+1)}{5n+6}=\frac{n+1}{5n+6}\)
LH
0
14 tháng 12 2023
5n+5n.52=650
5n(1+52)=650
5n.26=650
=>5n=650:26
=>5n=25=52
=>n=2
DT
1
TD
2
13 tháng 11 2016
Gọi d là \(ƯCLN\left(5n+2,5n+3\right)\)
\(\Rightarrow\begin{cases}5n+2⋮d\\5n+3⋮d\end{cases}\)
\(\Rightarrow\left(5n+3\right)-\left(5n+2\right)⋮d\)
\(\Rightarrow5n+3-5n-2⋮d\)
\(\Rightarrow1⋮d\Rightarrow d=1\RightarrowƯCLN\left(5n+2,5n+3\right)=1\)
Vậy 5n + 2 và 5n + 3 là hai số nguyên tố cùng nhau .
b, Gọi d là \(ƯCLN\left(7n+1,6n+1\right)\)
\(\Rightarrow\begin{cases}7n+1⋮d\\6n+1⋮d\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}42n+6⋮d\\42n+7⋮d\end{cases}\)
\(\Rightarrow\left(42n+7\right)-\left(42n+6\right)⋮d\)
\(\Rightarrow42n+7-42n-6⋮d\)
\(\Rightarrow1⋮d\Rightarrow d=1\RightarrowƯCLN\left(7n+1,6n+1\right)=1\)
Vậy 7n + 1 và 6n + 1 là hai số nguyên tố cùng nhau .
c, Gọi d là \(ƯCLN\left(5n+1,4n+1\right)\)
\(\Rightarrow\begin{cases}5n+1⋮d\\4n+1⋮d\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}20n+4⋮d\\20n+5⋮d\end{cases}\)
\(\Rightarrow\left(20n+5\right)-\left(20n+4\right)⋮d\)
\(\Rightarrow20n+5-20n-4⋮d\)
\(\Rightarrow1⋮d\Rightarrow d=1\RightarrowƯCLN\left(5n+1,4n+1\right)=1\)
Vậy 5n + 1 và 4n + 1 là hai số nguyên tố cùng nhau
5 tháng 4 2015
\(5A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{\left(5n+1\right).\left(5n+6\right)}\)
\(5A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{5n+1}-\frac{1}{5n+6}\)
\(5A=1-\frac{1}{5n+6}=\frac{5n+6-1}{5n+6}=\frac{5n+5}{5n+6}\)=> \(A=\frac{n+1}{5n+6}\)
PN
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 8 2023
CM: \(\dfrac{1}{1.6}\)+ \(\dfrac{1}{11.16}\)+...+ \(\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}\) = \(\dfrac{n+1}{5n+6}\)
A = \(\dfrac{1}{5}\)(\(\dfrac{5}{1.6}\) + \(\dfrac{5}{6.11}\)+...+ \(\dfrac{5}{\left(5n+1\right).\left(5n+6\right)}\))
A = \(\dfrac{1}{5}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{6}\)+ \(\dfrac{1}{6}\) - \(\dfrac{1}{11}\)+...+ \(\dfrac{1}{5n+1}\) - \(\dfrac{1}{5n+6}\))
A = \(\dfrac{1}{5}\) .( \(\dfrac{1}{1}\) - \(\dfrac{1}{5n+6}\))
A = \(\dfrac{1}{5}\). \(\dfrac{5n+6-1}{5n+6}\)
A = \(\dfrac{1}{5}\). \(\dfrac{5n+5}{5n+6}\)
A = \(\dfrac{1}{5}\) . \(\dfrac{5.\left(n+1\right)}{5n+6}\)
A = \(\dfrac{n+1}{5n+6}\)
⇒\(\dfrac{1}{1.6}\) + \(\dfrac{1}{6.11}\)+ \(\dfrac{1}{11.16}\)+...+ \(\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}\) = \(\dfrac{n+1}{5n+1}\) (đpcm)
8 tháng 8 2023
\(A=\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+...+\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}\)
\(A=\dfrac{1}{5}\left[1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right]\)
\(A=\dfrac{1}{5}\left(1-\dfrac{1}{5n+6}\right)\)
\(A=\dfrac{1}{5}\left(\dfrac{5n+6-1}{5n+6}\right)=\dfrac{1}{5}\left(\dfrac{5n+5}{5n+6}\right)=\dfrac{1}{5}.5\left(\dfrac{n+1}{5n+6}\right)=\dfrac{n+1}{5n+6}\)
\(\Rightarrow dpcm\)
NK
4
NL
29 tháng 4 2016
Gọi A = 1/1.6 + 1/6.11 +...+ 1/(5n+1)(5n+6)
5A = 5/1.6 + 5/6.11 + ... + 5/(5n+1)(5n+6)
=1 - 1/6 + 1/6 - 1/11 + ... + 1/5n+1 - 1/5n+6
=1 - 1/5n+6 =5n+6/5n+6 - 1/5n+6=5n+5 /5n+6
`5^n+5^(n+1)=30`
`=>5^n+5^n*5=30`
`=>5^n*(1+5)=30`
`=>5^n*6=30`
`=>5^n=30:6`
`=>5^n=5`
`=>5^n=5^1`
`=>n=1`
\(5^n+5^{n+1}=30\)
\(\Rightarrow5^n+5^n+5=30\)
\(\Rightarrow5^n+5^n=30-5\)
\(\Rightarrow5^n+5^n=25\)
\(\Rightarrow5^n+5^n=5^2\)
\(\Rightarrow n+n=2\)
\(\Rightarrow2n=2\)
\(\Rightarrow n=2:2\)
\(\Rightarrow n=1\)