a.Ta có: x+2xy-y=7
⇌ 2x+4xy-2y=14
⇌ 2x(2y+1) - (2y+1)=13
⇌ (2y+1)(2x-1) = 13
Do x,y ∈ Z+⇒ 2y+1;2x-1∈ Z+
Và (2y+1)(2x-1) = 13
Do 2y+1>2x-1 với mọi x,y ∈ Z+
⇒\(\left\{{}\begin{matrix}2x-1=1\\2y+1=13\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}2x=2\\2y=12\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x=1\text{(t/m)}\\y=6\text{(t/m)}\end{matrix}\right.\)
Vậy (x,y)∈\(\left\{\left(1,6\right)\right\}\)
b. Ta có: 2x+2y=2x+y
⇌ 2x- 2x+y+2y=0
⇌ -2x(2y-1)+(2y-1)=0
⇌ (2y-1)(1-2x)=0
⇌ \(\left[{}\begin{matrix}2^y-1=0\\1-2^x=0\end{matrix}\right.\) ⇒\(\left[{}\begin{matrix}2^y=1\\-2^x=-1\end{matrix}\right.\text{⇒}\left[{}\begin{matrix}y=0\text{(L)}\\x=0\text{(L)}\end{matrix}\right.\)
Vậy không tìm được các cặp (x,y) thỏa mãn bài toán