a. \(\left(x+8\right)⋮\left(x+4\right)\)
\(\Rightarrow\left(x+4\right)+4⋮\left(x+4\right)\)
Mà \(\left(x+4\right)⋮\left(x+4\right)\)
\(\Rightarrow4⋮\left(x+4\right)\)
\(\Rightarrow x+4\in\text{Ư} \left(4\right)=\left\{1;2;4\right\}\)
Ta có 3 trường hợp :
TH1 : \(x+4=1\Rightarrow x\notin N\) ( Loại )
TH2 : \(x+4=2\Rightarrow x\notin N\)(Loại )
TH3 : \(x+4=4\Rightarrow x=0\)
Vậy x = 0
a,Vì : \(x+8⋮x+2\)
Mà : \(x+2⋮x+2\)
\(\Rightarrow\left(x+8\right)-\left(x+2\right)⋮x+2\Rightarrow x+8-x-2⋮x+2\)
\(\Rightarrow6⋮x+2\Rightarrow x+2\inƯ\left(6\right)\)
Mà : \(Ư\left(6\right)=\left\{1;2;3;6\right\}\) ; \(x+2\ge2\Rightarrow x+2\in\left\{2;3;6\right\}\)
\(\Rightarrow x\in\left\{0;1;4\right\}\)
Vậy ...
b,Ta có : \(2y+7⋮y-1\) ; \(y-1⋮y-1\Rightarrow2\left(y-1\right)⋮y-1\Rightarrow2y-2⋮y-1\)
\(\Rightarrow\left(2y+7\right)-\left(2y-2\right)⋮y-1\Rightarrow2y+7-2y+2⋮y-1\)
\(\Rightarrow9⋮y-1\Rightarrow y-1\in\left\{1;3;9\right\}\Rightarrow y\in\left\{2;4;10\right\}\)
Vậy ...
c, Vì : \(x\in N\Rightarrow x-5\in N\)
\(y\in N\Rightarrow y+3\in N\left(y+3\ge3\right)\)
\(\Rightarrow x-5,y+3\inƯ\left(7\right)\)
Mà : \(Ư\left(7\right)=\left\{1;7\right\};y+3\ge3\)
\(\Rightarrow x-5=1\Rightarrow x=6;y+3=7\Rightarrow y=4\)
Vậy ...
d,Vì : \(\overline{3x7y}⋮2,5\Rightarrow y=0\)
Ta có : \(\overline{3x70}⋮\) 9 dư 1
\(\Rightarrow3+x+7+0⋮\) 9 dư 1
\(\Rightarrow10+x⋮\) 9 dư 1 \(\Rightarrow9+x⋮9\)
Vì : \(9⋮9\Rightarrow x⋮9\Rightarrow x\in\left\{0;9\right\}\)
Vậy x = 0 thì b = 0
x = 9 thì b = 0
e, Vì : \(ƯCLN\left(x,y\right)=9\Rightarrow\begin{cases}x=9.k_1\\y=9.k_2\end{cases}\) với ƯCLN(k1,k2) = 1
Thay vào x + y = 63 ta có :
\(9.k_1+9.k_2=63\Rightarrow9\left(k_1+k_2\right)=63\Rightarrow k_1+k_2=7\)
Mà : x > y > 9
+) Nếu : k1 = 6 ; k2 = 1 \(\Rightarrow\begin{cases}x=6.9=54\\y=1.9=9\end{cases}\) ( loại )
+) Nếu : k1 = 5 ; k2 = 2 \(\Rightarrow\begin{cases}x=5.9=45\\y=2.9=18\end{cases}\) ( nhận )
+) Nếu : k1 = 4 ; k2 = 3 \(\Rightarrow\begin{cases}x=9.4=36\\y=9.3=27\end{cases}\) ( nhận )
Vậy x = 45 thì y = 18
x = 36 thì y = 27
