a, Ta có: 3xy - 5 = x2 + 2y
=> 3xy - x2 - 2y = 5
=> y.( 3x - 2 ) = 5 + x.x
=> y = \(\frac{5+x^2}{3x-2}\)
=> \(x^2+5⋮3x-2\)( vì y là số nguyên )
=> \(3x^2+15⋮3x-2\)
\(\Rightarrow x\left(3x-2\right)+15+2x⋮3x-2\)
\(\Rightarrow2x+15⋮3x+2\)
\(\Rightarrow6x+45⋮3x+2\)
\(\Rightarrow2.\left(3x+2\right)+41⋮3x+2\)
\(\Rightarrow41⋮3x+2\)
\(\Rightarrow3x+2\in\left\{-41;-1;1;41\right\}\)
\(\Rightarrow3x\in\left\{-43;-3;-1;39\right\}\)
VÌ 3x chia hết cho 3
\(\Rightarrow3x\in\left\{-3;39\right\}\)
\(\Rightarrow x\in\left\{-1;13\right\}\)
+) với x = -1 => y = -6/5 ( loại )
+) với x = 13 => y = 174/37 ( loại )
Vậy không tìm được ( x ; y ) thỏa mãn bài
b,
Xét \(3^{n+2}-2^{n+2}+3^n-2^n=3^n.9-2^n.4+3^n-2^n=3^n.\left(9+1\right)-2^n.\left(4+1\right)=3^n.10-2^n.5\)
\(=3^n.10-2^{n-1}.2.5=3^n.10-2^{n-1}.10=10.\left(3^n-2^{n-1}\right)⋮10\)
\(\Rightarrow3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
Vậy: \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
