Tìm các số nguyên x,y biết (x-3)(y-3) =25 Gíúp mình vs :(( 28/08/2021 Bởi Mary Tìm các số nguyên x,y biết (x-3)(y-3) =25 Gíúp mình vs :((
Đáp án: Giải thích các bước giải: Lập bảng số nguyên ta có : $\begin{array}{|c|c|}\hline x-3&25&1&-1&-25&5&-5\\\hline x&28&4&2&-22&8&-2\\\hline\end{array}$ $\begin{array}{|c|c|}\hline y-3&1&25&-25&-1&5&-5\\\hline y&4&28&-22&2&8&-2\\\hline\end{array}$ ta có: `(x;y) = (28;4); (4;28); (-2;-22); (-22;-2); (8;8); (-2;-2)` Bình luận
*Lời giải : `(x -3) (y – 3) = 25` `-> x – 3, y -3 = 25 . 1 = 1 . 25 = (-1) . (-25) = (-25) . (-1) = 5 . 5 = (-5) . (-5)` `*` \(\left\{ \begin{array}{l}x-3=25\\y-3=1\end{array} \right.\) `⇔` \(\left\{ \begin{array}{l}x=28\\y=4\end{array} \right.\) `*` \(\left\{ \begin{array}{l}x-3=1\\y-3=25\end{array} \right.\) `⇔` \(\left\{ \begin{array}{l}x=4\\y=28\end{array} \right.\) `*` \(\left\{ \begin{array}{l}x-3=-1\\y-3=-25\end{array} \right.\) `⇔` \(\left\{ \begin{array}{l}x=2\\y=-22\end{array} \right.\) `*` \(\left\{ \begin{array}{l}x-3=-25\\y-3=-1\end{array} \right.\) `⇔` \(\left\{ \begin{array}{l}x=-22\\y=2\end{array} \right.\) `*` \(\left\{ \begin{array}{l}x-3=5\\y-3=5\end{array} \right.\) `⇔` \(\left\{ \begin{array}{l}x=8\\y=8\end{array} \right.\) `*` \(\left\{ \begin{array}{l}x-3=-5\\y-3=-5\end{array} \right.\) `⇔` \(\left\{ \begin{array}{l}x=-2\\y=-2\end{array} \right.\) Vậy `(x;y) = (28;4); (4;28); (2;-22); (-22;2); (8;8); (-2;-2)` Bình luận
Đáp án:
Giải thích các bước giải:
Lập bảng số nguyên ta có :
$\begin{array}{|c|c|}\hline x-3&25&1&-1&-25&5&-5\\\hline x&28&4&2&-22&8&-2\\\hline\end{array}$
$\begin{array}{|c|c|}\hline y-3&1&25&-25&-1&5&-5\\\hline y&4&28&-22&2&8&-2\\\hline\end{array}$
ta có:
`(x;y) = (28;4); (4;28); (-2;-22); (-22;-2); (8;8); (-2;-2)`
*Lời giải :
`(x -3) (y – 3) = 25`
`-> x – 3, y -3 = 25 . 1 = 1 . 25 = (-1) . (-25) = (-25) . (-1) = 5 . 5 = (-5) . (-5)`
`*` \(\left\{ \begin{array}{l}x-3=25\\y-3=1\end{array} \right.\) `⇔` \(\left\{ \begin{array}{l}x=28\\y=4\end{array} \right.\)
`*` \(\left\{ \begin{array}{l}x-3=1\\y-3=25\end{array} \right.\) `⇔` \(\left\{ \begin{array}{l}x=4\\y=28\end{array} \right.\)
`*` \(\left\{ \begin{array}{l}x-3=-1\\y-3=-25\end{array} \right.\) `⇔` \(\left\{ \begin{array}{l}x=2\\y=-22\end{array} \right.\)
`*` \(\left\{ \begin{array}{l}x-3=-25\\y-3=-1\end{array} \right.\) `⇔` \(\left\{ \begin{array}{l}x=-22\\y=2\end{array} \right.\)
`*` \(\left\{ \begin{array}{l}x-3=5\\y-3=5\end{array} \right.\) `⇔` \(\left\{ \begin{array}{l}x=8\\y=8\end{array} \right.\)
`*` \(\left\{ \begin{array}{l}x-3=-5\\y-3=-5\end{array} \right.\) `⇔` \(\left\{ \begin{array}{l}x=-2\\y=-2\end{array} \right.\)
Vậy `(x;y) = (28;4); (4;28); (2;-22); (-22;2); (8;8); (-2;-2)`