Câu hỏi của jiyootrann

Question

Câu 1: Tìm các cặp số tự nhiên x ; y biết :

a) (3x - 2)

Nguyễn Hoàng Minh · Accepted Answer

$1,\ a,\left(3x-2\right)\left(2y-3\right)=1\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x-2=1\2y-3=1\end{matrix}\right.\\left\{{}\begin{matrix}3x-2=-1\2y-3=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\y=2\end{matrix}\right.\\left\{{}\begin{matrix}x=\dfrac{1}{3}\y=1\end{matrix}\right.\end{matrix}\right.\ \Leftrightarrow\left(x;y\right)=\left\{\left(1;2\right);\left(\dfrac{1}{3};1\right)\right\}$

$b,\left(2x+1\right)\left(y-3\right)=10$

Ta có bảng

$2x+1$	1	2	5	10	$-1$	$-2$	$-5$	$-10$
$y-3$	$10$	$5$	$2$	$1$	$-10$	$-5$	$-2$	$-1$
$x$	1	$\dfrac{1}{2}$	2	$\dfrac{9}{2}$	$-1$	$-\dfrac{3}{2}$	$-3$	$-\dfrac{11}{2}$
$y$	13	8	5	4	$-7$	$-2$	1	2

Vậy $\left(x;y\right)=...$

Câu trả lời:

$1,\ a,\left(3x-2\right)\left(2y-3\right)=1\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x-2=1\2y-3=1\end{matrix}\right.\\left\{{}\begin{matrix}3x-2=-1\2y-3=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\y=2\end{matrix}\right.\\left\{{}\begin{matrix}x=\dfrac{1}{3}\y=1\end{matrix}\right.\end{matrix}\right.\ \Leftrightarrow\left(x;y\right)=\left\{\left(1;2\right);\left(\dfrac{1}{3};1\right)\right\}$

$b,\left(2x+1\right)\left(y-3\right)=10$

Ta có bảng

$2x+1$	1	2	5	10	$-1$	$-2$	$-5$	$-10$
$y-3$	$10$	$5$	$2$	$1$	$-10$	$-5$	$-2$	$-1$
$x$	1	$\dfrac{1}{2}$	2	$\dfrac{9}{2}$	$-1$	$-\dfrac{3}{2}$	$-3$	$-\dfrac{11}{2}$
$y$	13	8	5	4	$-7$	$-2$	1	2

Vậy $\left(x;y\right)=...$

\(2x+1\)	1	2	5	10	\(-1\)	\(-2\)	\(-5\)	\(-10\)
\(y-3\)	\(10\)	\(5\)	\(2\)	\(1\)	\(-10\)	\(-5\)	\(-2\)	\(-1\)
\(x\)	1	\(\dfrac{1}{2}\)	2	\(\dfrac{9}{2}\)	\(-1\)	\(-\dfrac{3}{2}\)	\(-3\)	\(-\dfrac{11}{2}\)
\(y\)	13	8	5	4	\(-7\)	\(-2\)	1	2

Chúng tôi đề xuất

Tài nguyên

Ứng dụng mobile

Bảng xếp hạng

Các khóa học có thể bạn quan tâm

Yêu cầu VIP