Question

Tìm x y biết

a)xy+3x-2y=11

b)2x^2-2xy+x-y=12

c)2xy-10y-x=13

e)xy-2y^2+8y-3x=13

f)xy-2y^2+8y-3x=13

Answer 1

\(a)xy+3x-2y=11\)

\(\Leftrightarrow xy+3x-2y-6=5\)

\(\Leftrightarrow x\left(y+3\right)-2\left(y+3\right)=5\)

\(\Leftrightarrow\left(y+3\right)\left(x-2\right)=5\)

\(\Leftrightarrow\hept{\begin{cases}y+3=-1\\x-2=-5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-4\\x=-3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y+3=1\\x-2=5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-2\\x=7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y+3=-5\\x-2=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-8\\x=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y+3=5\\x-2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}y=2\\x=3\end{cases}}\)

Answer 2

\(b)2x^2-2xy+x-y=12\)

\(\Leftrightarrow2x\left(x-y\right)+\left(x-y\right)=12\)

\(\Leftrightarrow\left(x-y\right)\left(2x+1\right)=12\)

\(\Rightarrow\left(x-y\right);\left(2x+1\right)\inƯ\left(12\right)\)

\(\RightarrowƯ\left(12\right)\in\left\{-1;1;-2;2;-3;3;-4;4;-6;6;-12;12\right\}\)

Vì 2x+1 luôn lẻ

\(\Rightarrow2x+1\in\left\{-1;1;-3;3\right\}\)

\(\Leftrightarrow\hept{\begin{cases}2x+1=-1\\x-y=-12\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=11\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x+1=1\\x-y=12\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=-12\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x+1=-3\\x-y=-4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x+1=3\\x-y=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)

Answer 3

\(c)2xy-10y-x=13\)

\(\Leftrightarrow x\left(2y-1\right)-2y.5+5=18\)

\(\Leftrightarrow x\left(2y-1\right)-5\left(2y-1\right)=18\)

\(\Leftrightarrow\left(2y-1\right)\left(x-5\right)=18\)

\(\Leftrightarrow2y-1;x-5\inƯ\left(18\right)\)

\(\RightarrowƯ\left(18\right)\in\left\{-1;1;-2;2;-3;3;-6;6;-9;9;-18;18\right\}\)

Vì 2y-1  luôn lẻ

=>2y-1 thuộc {-1;1;-3;3;-9;9}

=> Làm  tương tự nhé

\(e)xy-2y^2+8y-3x=13\)

\(\Leftrightarrow xy-2y^2+2y+6y-3x-6=7\)

\(\Leftrightarrow y\left(x-2y+2\right)+3\left(-x+2y-2\right)=7\)

\(\Leftrightarrow y\left(x-2y+2\right)-3\left(x-2y+2\right)=7\)

\(\Leftrightarrow\left(x-2y+2\right)\left(y-3\right)=7\)

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