(x – 2) (y – 3) = 0 (x – 2) (y – 5) = 2 (x – 5) (y – 6) = 11 với x,y thuộc Z 12/09/2021 Bởi Delilah (x – 2) (y – 3) = 0 (x – 2) (y – 5) = 2 (x – 5) (y – 6) = 11 với x,y thuộc Z
Đáp án: Giải thích các bước giải: ta có: (x-2)(y-3)=0 ⇒\(\left[ \begin{array}{l}x-2=0\\y-3=0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=2\\y=3\end{array} \right.\) (x-2)(y-5)=0 ⇒\(\left[ \begin{array}{l}x-2=2\\y-5=1\end{array} \right.\) \(\left[ \begin{array}{l}x-2=1\\y-5=2\end{array} \right.\) \(\left[ \begin{array}{l}x-2=-2\\y-5=-1\end{array} \right.\) \(\left[ \begin{array}{l}x-2=-1\\y-5=-2\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=4\\y=6\end{array} \right.\) \(\left[ \begin{array}{l}x=3\\y=7\end{array} \right.\) \(\left[ \begin{array}{l}x=0\\y=4\end{array} \right.\) \(\left[ \begin{array}{l}x=1\\y=3\end{array} \right.\) (x-5)(y-6)=1 ⇒\(\left[ \begin{array}{l}x-5=11\\y-6=1\end{array} \right.\) \(\left[ \begin{array}{l}x-5=1\\y-6=11\end{array} \right.\) \(\left[ \begin{array}{l}x-5=-11\\y-6=-1\end{array} \right.\) \(\left[ \begin{array}{l}x-5=-1\\y-6=-11\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=16\\y=7\end{array} \right.\) \(\left[ \begin{array}{l}x=6\\y=17\end{array} \right.\) \(\left[ \begin{array}{l}x=-6\\y=5\end{array} \right.\) \(\left[ \begin{array}{l}x=4\\y=-5\end{array} \right.\) Bình luận
Đáp án + giải thích bước giải : $1/$ `(x – 2) (y – 3) = 0` `⇔` \(\left[ \begin{array}{l}x-2=0\\y-3=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=0+2\\y=0+3\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=2\\y=3\end{array} \right.\) Vậy `(x;y) ∈ {2;3}` thỏa mãn `x,y ∈ ZZ` $2/$ `(x – 2) (y – 5) = 2` `⇔` \(\left\{ \begin{array}{l}x-2\\y-5\end{array} \right.\) `∈ Ư (2) = {±1; ±2}` `⇔` \(\left\{ \begin{array}{l}x=3\\x=1\\x=4\\x=0\end{array} \right.\) `⇔` \(\left\{ \begin{array}{l}y=6\\y=4\\y=7\\y=3\end{array} \right.\) Vậy `x ∈ {3;1;4;0}; y ∈ {6;4;7;3}` thỏa mãn `x,y ∈ ZZ` $3/$ `(x – 5) (y – 6) = 11` `⇔` \(\left\{ \begin{array}{l}x-5\\y-6\end{array} \right.\) `∈ Ư (11) = {±1 ;±11}` `⇔` \(\left\{ \begin{array}{l}x=6\\x=4\\x=16\\x=-6\end{array} \right.\) `⇔` \(\left\{ \begin{array}{l}y=7\\y=5\\y=17\\y=-5\end{array} \right.\) Vậy `x ∈ {6;4;16;-6}; y ∈ {7;5;17;-5}` thỏa mãn `x,y ∈ ZZ` Bình luận
Đáp án:
Giải thích các bước giải:
ta có:
(x-2)(y-3)=0
⇒\(\left[ \begin{array}{l}x-2=0\\y-3=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=2\\y=3\end{array} \right.\)
(x-2)(y-5)=0
⇒\(\left[ \begin{array}{l}x-2=2\\y-5=1\end{array} \right.\)
\(\left[ \begin{array}{l}x-2=1\\y-5=2\end{array} \right.\)
\(\left[ \begin{array}{l}x-2=-2\\y-5=-1\end{array} \right.\)
\(\left[ \begin{array}{l}x-2=-1\\y-5=-2\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=4\\y=6\end{array} \right.\)
\(\left[ \begin{array}{l}x=3\\y=7\end{array} \right.\)
\(\left[ \begin{array}{l}x=0\\y=4\end{array} \right.\)
\(\left[ \begin{array}{l}x=1\\y=3\end{array} \right.\)
(x-5)(y-6)=1
⇒\(\left[ \begin{array}{l}x-5=11\\y-6=1\end{array} \right.\)
\(\left[ \begin{array}{l}x-5=1\\y-6=11\end{array} \right.\)
\(\left[ \begin{array}{l}x-5=-11\\y-6=-1\end{array} \right.\)
\(\left[ \begin{array}{l}x-5=-1\\y-6=-11\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=16\\y=7\end{array} \right.\)
\(\left[ \begin{array}{l}x=6\\y=17\end{array} \right.\)
\(\left[ \begin{array}{l}x=-6\\y=5\end{array} \right.\)
\(\left[ \begin{array}{l}x=4\\y=-5\end{array} \right.\)
Đáp án + giải thích bước giải :
$1/$ `(x – 2) (y – 3) = 0`
`⇔` \(\left[ \begin{array}{l}x-2=0\\y-3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0+2\\y=0+3\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=2\\y=3\end{array} \right.\)
Vậy `(x;y) ∈ {2;3}` thỏa mãn `x,y ∈ ZZ`
$2/$ `(x – 2) (y – 5) = 2`
`⇔` \(\left\{ \begin{array}{l}x-2\\y-5\end{array} \right.\) `∈ Ư (2) = {±1; ±2}`
`⇔` \(\left\{ \begin{array}{l}x=3\\x=1\\x=4\\x=0\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}y=6\\y=4\\y=7\\y=3\end{array} \right.\)
Vậy `x ∈ {3;1;4;0}; y ∈ {6;4;7;3}` thỏa mãn `x,y ∈ ZZ`
$3/$ `(x – 5) (y – 6) = 11`
`⇔` \(\left\{ \begin{array}{l}x-5\\y-6\end{array} \right.\) `∈ Ư (11) = {±1 ;±11}`
`⇔` \(\left\{ \begin{array}{l}x=6\\x=4\\x=16\\x=-6\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}y=7\\y=5\\y=17\\y=-5\end{array} \right.\)
Vậy `x ∈ {6;4;16;-6}; y ∈ {7;5;17;-5}` thỏa mãn `x,y ∈ ZZ`