\(\frac{n-6}{n-4}\)
tim n \(\in\)Z
giup minh voi ,minh dang can gap
Tim cac so nguyen m,n,p thoa man p = m - n = m . n
Giup minh voi minh dang can gap
Ta có : \(m-n=mn\)
\(\Leftrightarrow mn-m+n=0\)
\(\Leftrightarrow m\left(n-1\right)+n-1=-1\)
\(\Leftrightarrow\left(m+1\right)\left(n-1\right)=-1\)\(\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}m+1=-1\\n-1=1\end{cases}}\\\hept{\begin{cases}m+1=1\\n-1=-1\end{cases}}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}m+1=-1\\n-1=1\end{cases}}\\\hept{\begin{cases}m+1=1\\n-1=-1\end{cases}}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}m+1=1\\n-1=-1\end{cases}}\\\hept{\begin{cases}m+1=-1\\n-1=1\end{cases}}\end{cases}}\)
tim so tu nhien n de n +3 chia het cho n-1
gup minh voi dang can gap
n+3 chia hết cho n-1
n-1 chia hết cho n-1
nên n+3-(n-1) chia hết cho n-1
<=> n+3 - (n-1) = n+3- n+1 = 4
<=> 4 chia hết cho n-1
vậy n-1 thuộc ước của 4
n-1 =2 => n=3
n-1 =1 => n=2
n-1 =4 => n=5
De n+3 chia het cho n-1
n+3=n-1+4 chia het cho n-1
vi n-1 chia het cho n-1 nen de n-1+4 chia het cho n-1 thi 4 phai chia het cho n-1
suy ra n-1 thuoc U(4)={1;2;4}
ta co bang sau:
n-1 1 2 4
n 2 3 5
ban thong cam minh ko biet ve hinh
Tim n€Z , biet:
a) n+8/n-5 la so nguyen.
b)n2+1/n-6 la so nguyen.
Giup minh voi minh dang can gap .
Lam chi tiet ho minh nha.
Q= 1/2! + 2/3! + 3/4! + ... + n-1/n!
Chung minh rang : Q<1
Minh dang can gap. Giup minh voi!!
Tim x :
x/2 + x/6 + x/12+ ................................................ + x/90 = 9
giuy minh voi minh dang can gap
ta có: x/1.2+x/2.3+x/3.4+.....+x/9.10=9
x-x/2+x/2-x/3+x/3-......+x/9-x/10=9
x-x/10=9
=>x=10
Tim chu so tan cung cua A=2^0+2^2+2^4+2^6+...+2^2014+2^2016
Tim so tu nhien n biet 82 chia n du 7 va 144 chia n du 9
MINH DANG CAN GAP CAC BAN GIAI NHANH GIUM CHO MAI MINH PHAI NOP ROI
tim mot phan so biet rang neu phan so do tru di 1/5 roi lai them 1/4 thi duoc 5/4. Phan so can tim la: .................................
Giup minh voi minh dang can gap
tim 4 so tan cung cua 321978
GIUP MINH VOI MINH DANG CAN GAP
Ta thấy :1978:4=494 dư 2
Ta có:321978 = 32(494)4 .32.2
=*******24 .***4
=*********6.***4
=*******4
Các *** để biểu thị cho các số,vì dài qá k viết đc hết ra
giup minh voi nha
minh dang can gap
(3n+2)4=( 3n+2)6
Ta có :
\(\left(3n+2\right)^4=\left(3n+2\right)^6\)
\(\Leftrightarrow\left(3n+2\right)^6-\left(3n+2\right)^4=0\)
\(\Leftrightarrow\left[\left(3n+2\right)^2-1\right]-\left(3n+4\right)^4=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[\left(3n+2\right)^2-1\right]=0\\\left(3n+2\right)^4=0\end{matrix}\right.\)
+)\(\left(3n+2\right)^4=0\)
\(\Leftrightarrow n=\dfrac{2}{3}\)\(\left(tm\right)\)
+) \(\left[\left(3n+2\right)^2-1\right]=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3n+2=1\\3n+2=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}n=\dfrac{-1}{3}\\n=-1\end{matrix}\right.\)\(\left(tm\right)\)
Vậy ....................
Ta có: \(\left(3n+2\right)^4=\left(3n+2\right)^6\)
\(\Leftrightarrow\left[{}\begin{matrix}3n+2=1\\3n+2=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}3n=1-2\\3n=-1-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}3n=-1\\3n=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}n=\dfrac{-1}{3}\\n=-1\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}n=\dfrac{-1}{3}\\n=-1\end{matrix}\right.\)