2x-xy-3y+1=0
tìm x, y
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a, Vì |2x+8| và |3y-9x| đều >= 0
=> |2x+8| + |3y-9x| >= 0
Dấu "=" xảy ra <=> 2x+8=0 và 3y-9x=0 <=> x=-4 và y=-12
Vậy x=-4 và y=-12
Tk mk nha
a.
\(\left\{{}\begin{matrix}\left(x-1\right)^2-\left(y+1\right)^2=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1-y-1\right)\left(x-1+y+1\right)=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-2\right)\left(x+y\right)=0\\x+3y-5=0\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x-y-2=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{4}\\y=\dfrac{3}{4}\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+y=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)
b.
\(\left\{{}\begin{matrix}xy-2x-y+2=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-2\right)-\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
TH1:
\(\left\{{}\begin{matrix}x-1=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)
TH2:
\(\left\{{}\begin{matrix}y-2=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
2x-3y=0\(\Rightarrow2x=3y\Rightarrow x=\frac{3}{2}y\)
Mà xy-150=0
Hay \(\frac{3}{2}y\cdot y=150\)
\(y^2=150:\frac{3}{2}\)
\(y^2=100\)
y=10 hoặc y=-10
Nếu y=10\(\Rightarrow x=\frac{3}{2}\cdot10=15\)
Nếu y=-10\(\Rightarrow x=\frac{3}{2}\cdot\left(-10\right)=-15\)
suy ra x.(y-2)-3.(y-2)+6+1=0
suy ra (x-3).(y-2)=-7
suy ra x-3;y-2 thuộc Ư(-7)
tự lập bảng tự tính
\(2x-xy-3y+1=0\)
\(\Rightarrow2x-xy-3y+1+5=5\)
\(\Rightarrow2x-xy-3y+6=5\)
\(\Rightarrow x\left(2-y\right)+3\left(2-y\right)=5\)
\(\Rightarrow\left(x+3\right)\left(2-y\right)=5\)
\(\Rightarrow x+3;2-y\inƯ\left(5\right)\)
\(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3=1\Rightarrow x=-2\\2-y=5\Rightarrow y=-3\end{matrix}\right.\\\left\{{}\begin{matrix}x+3=-1\Rightarrow x=-4\\2-y=-5\Rightarrow y=7\end{matrix}\right.\\\left\{{}\begin{matrix}x+3=5\Rightarrow x=2\\2-y=1\Rightarrow y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x+3=-5\Rightarrow x=-8\\2-y=-1\Rightarrow y=3\end{matrix}\right.\end{matrix}\right.\)
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