Toán Lớp 7: Tìm các số nguyên x,y bt
x-2xy+y-3=0
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Toán Lớp 7: Tìm các số nguyên x,y bt
x-2xy+y-3=0
Comments ( 2 )
x-2xy + y-3=0
->2x-4xy+2y-6=0
->2x(1-2y) + 2y-1 = 5
-> 2x(1-2y) – 1(1-2y)=5
->(2x-1)(1-2y)=1.5=5.1=(-1).(-5)=(-5).(-1)
TH1 :
2x-1=1, 1-2y=5
->x=1, y=-2 (Tm)
TH2 :
2x-1=5,1-2y=1
->x=3,y=0 (Tm)
TH3 :
2x-1=-1,1-2y=-5
->x=0, y=3 (Tm)
TH4 :
2x-1=-5, 1-2y=-1
->x=-2, y=1 (Tm)
Vậy (x;y)=(1;-2), (3;0), (0;3), (-2;1)
Giải đáp:
$\left( {x;y} \right) \in \left\{ {\left( {1; – 2} \right);\left( { – 2;1} \right);\left( {0;3} \right);\left( {3;0} \right)} \right\}$
Lời giải và giải thích chi tiết:
Ta có:
$\begin{array}{l}
x – 2xy + y – 3 = 0\\
\Leftrightarrow 2x – 4xy + 2y – 6 = 0\\
\Leftrightarrow 2x\left( {1 – 2y} \right) + \left( {2y – 1} \right) – 5 = 0\\
\Leftrightarrow \left( {2y – 1} \right)\left( {1 – 2x} \right) = 5 (1)
\end{array}$
Do $x,y\in Z$ nên $1-2x,2y-1$ là ước nguyên của $5$
Khi đó:
$\left( 1 \right) \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2y – 1 = 1\\
1 – 2x = 5
\end{array} \right.\\
\left\{ \begin{array}{l}
2y – 1 = – 1\\
1 – 2x = – 5
\end{array} \right.\\
\left\{ \begin{array}{l}
2y – 1 = 5\\
1 – 2x = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
2y – 1 = – 5\\
1 – 2x = – 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
y = 1\\
x = – 2
\end{array} \right.\\
\left\{ \begin{array}{l}
y = 0\\
x = 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 3\\
y = 0
\end{array} \right.\\
\left\{ \begin{array}{l}
y = – 2\\
x = 1
\end{array} \right.
\end{array} \right.$
Vậy $\left( {x;y} \right) \in \left\{ {\left( {1; – 2} \right);\left( { – 2;1} \right);\left( {0;3} \right);\left( {3;0} \right)} \right\}$