\(10^x+39=5y\)
\(5y-10^x=39\)
Vì \(Ư\left(39\right)\in\left\{1;3;13;39\right\}\)
\(\Rightarrow x=0\)
Thế x = 0, ta có:
\(5y-1=39\)
\(5y=39+1\)
\(5y=40\)
\(y=40:5\)
\(y=8\)
Vậy:\(x=0,y=8\)
\(10^x\)+39=5y
5y−\(10^x\)=39
Vì Ư(39)∈{1;3;13;39}
⇒x=0
Thế x = 0, ta có:
5y−1=39
5y=39+1
5y=40
y=40:5
y=8