2 câu trả lời
Đáp án:
`(x ;y) = (3 ; -1 ) ;(1 ; 3) ; (7 ; 3) ; (-3 ; -1)`
Giải thích các bước giải:
`x^2 -2xy -2x+4y =5`
`=> (x^2 -2xy) -(2x -4y) =5`
`=> x(x -2y) - 2(x- 2y) =5`
`=> (x-2)(x-2y) =5`
`=> (x-2) ;(x-2y) \in Ư(5) ={\pm 1; \pm 5}`
Từ đó , ta có bảng sau :
$\begin{array}{|c|c|}\hline x-2&1&-1&5&-5 \\\hline x&3&1&7&-3 \\\hline x-2y&5&-5&1&-1 \\\hline y&-1&3&3&-1 \\\hline \end{array}$
Vậy `(x ;y) = (3 ; -1 ) ;(1 ; 3) ; (7 ; 3) ; (-3 ; -1)`
`x^2 − 2xy−2x+4y = 5`
`→x(x−2y)−2(x−2y)=5`
`→(x−2)(x−2y)=5 `
`→ (x−2);(x−2y) ∈ Ư(5)={ 1; 5}`
`TH1 : x−2=1`
`→x = 3x−2y = 5`
`→ 3−2y = 5`
`→2y=−2`
`→y=−1`
`TH2 : x−2=−1`
`→x=1x−2y=5`
`→1−2y=5 `
`→2y=−4`
`→y=−2 `
`TH3 : x−2=5`
`→x=7x−2y=1 `
`→7−2y=1`
`→2y=6`
`→y=3`
`TH4 : x−2=−5`
`→x=−3x−2y=1`
`→−3−2y=1`
`→2y=−4`
`→y=−2`