5n + 5n+2 = 650
3n-1+5x3n-1=162
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JE
1
10 tháng 2 2021
a,\(lim\dfrac{1-2n^2}{5n+5}=lim\dfrac{\left(1-n\sqrt{2}\right)\left(1+n\sqrt{2}\right)}{5n+5}=lim\dfrac{\left(\dfrac{1}{n}-\sqrt{2}\right)\left(\dfrac{1}{n}+\sqrt{2}\right)}{5+\dfrac{5}{n}}=\dfrac{-2}{5}\)
b,\(lim\dfrac{1-2n}{5n+5n^2}=lim\dfrac{\dfrac{1}{n^2}-\dfrac{2}{n}}{\dfrac{5}{n}+5}=\dfrac{0}{5}=0\)
10 tháng 2 2021
a,\(lim\dfrac{n^2-2n}{5n+3n^2}=lim\dfrac{1-\dfrac{2}{n}}{\dfrac{5}{n}+3}=\dfrac{1}{3}\)
b,\(lim\dfrac{n^2-2}{5n+3n^2}=lim\dfrac{1-\dfrac{2}{n^2}}{\dfrac{5}{n}+3}=\dfrac{1}{3}\)
c,\(lim\dfrac{1-2n}{5n+3n^2}=lim\dfrac{1-2n}{n\left(5+3n\right)}=lim\dfrac{\dfrac{1}{n}-2}{1\left(\dfrac{5}{n}+3\right)}=-\dfrac{2}{3}\)
d,\(lim\dfrac{1-2n^2}{5n+5}=lim\dfrac{\left(1-n\sqrt{2}\right)\left(1+n\sqrt{2}\right)}{5n+5}=lim\dfrac{\left(\dfrac{1}{n}-\sqrt{2}\right)\left(\dfrac{1}{n}+\sqrt{2}\right)}{5+\dfrac{5}{n}}=\dfrac{-2}{5}\)
TU
6
NH
0
TH
1
9 tháng 12 2015
35n+2+35n+1-35n
=35n.32+35n.31-35n
=35n.9+35n.3-35n
=35n.(9+3-1)
=35n.11 chia hết cho 11
=> 35n+2+35n+1-35n chia hết cho 11
JG
2
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
26 tháng 10 2023
\(3^{5n+2}+3^{5n+1}-3^{5n}=3^{5n}\left(3^2+3-1\right)=11.3^{5n}⋮11\)
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
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
3n-1+5.3n-1=162
3n-1.(5+1)=162
3n-1.6=162
3n-1=162:6
3n-1=27=33
=>n-1=3
=>n=3+1
=>n=4
5n+5n+2=650
5n+5n.52=650
5n.(52+1)=650
5n.26=650
5n=650:26
5n=25=52
=>n=2