1 câu trả lời
Bạn tham khảo nhé.
Bổ sung là `x,y\inZZ` nhé.
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`3x^2+3xy-17=7x-2y`
`<=>3x^2+3xy-7x+2y-17=0`
`<=>3x(x+y)+2x+2y-9x-17=0`
`<=>3x(x+y)+2(x+y)-9x-17=0`
`<=>(x+y)(3x+2)-9x-6-11=0`
`<=>(x+y)(3x+2)-3(3x+2)=11`
`<=>(x+y-3)(3x+2)=11`
Vì `x,y\inZZ`
`=>(x+y-3)` và `(3x+2)` `\inƯ(11)\in{+-1;+-11}`
`\text{TH1}`
`{(x+y-3=1),(3x+2=11):}`
`<=>{(x+y=4),(3x=9):}`
`<=>{(y=4-x),(x=3):}`
`<=>{(y=4-3),(x=3):}`
`<=>{(y=1),(x=3):}(\text{Nhận})`
`\text{TH2}`
`{(x+y-3=-1),(3x+2=-11):}`
`<=>{(x+y=2),(3x=-13):}`
`<=>{(y=2-x),(x=-13/3):}(\text{Loại})`
`\text{TH3}`
`{(x+y-3=11),(3x+2=1):}`
`<=>{(x+y=14),(3x=-1):}`
`<=>{(y=14-x),(x=-1/3):}(\text{Loại})`
`\text{TH4}`
`{(x+y-3=-11),(3x+2=-1):}`
`<=>{(x+y=-8),(3x=-3):}`
`<=>{(y=-8-x),(x=-1):}`
`<=>{(y=-8-(-1)),(x=-1):}`
`<=>{(y=-7),(x=-1):}(\text{Nhận})`
Vậy `(x;y)\in(-1;-7),(3;1)`