tìm x,y thỏa mãn
a, |12x+8|+|11y-5| < 0
b, |3x+2y|+|4y-1|< 0
c,|x+y-7|+|xy-10|< 0
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a: \(\Leftrightarrow\left(x;y-3\right)\in\left\{\left(1;17\right);\left(17;1\right);\left(-1;-17\right);\left(-17;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(1;20\right);\left(17;4\right);\left(-1;-14\right);\left(-17;2\right)\right\}\)
b: \(\Leftrightarrow\left(x-1;y+2\right)\in\left\{\left(1;7\right);\left(7;1\right);\left(-1;-7\right);\left(-7;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(2;5\right);\left(8;-1\right);\left(0;-9\right);\left(-6;-3\right)\right\}\)
c: =>(y+1)(3x+1)=7
=>\(\left(3x+1;y+1\right)\in\left\{\left(1;7\right);\left(7;1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(0;6\right);\left(2;0\right)\right\}\)
Ta có: \(\left\{{}\begin{matrix}\left|12x+8\right|\ge0\\\left|11y-5\right|\ge0\\\left|13z-y+1\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|12x+8\right|+\left|11y-5\right|+\left|13z-y+1\right|\ge0\)
Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}\left|12x+8\right|=0\\\left|11y-5\right|=0\\\left|13z-y+1\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}12x+8=0\\11y-5=0\\13z-y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{8}{12}=-\dfrac{2}{3}\\y=\dfrac{5}{11}\\z=\dfrac{-1+y}{13}=\dfrac{-1+\dfrac{5}{11}}{13}=-\dfrac{6}{143}\end{matrix}\right.\)
Vậy.....................
Ta có: \(\left\{{}\begin{matrix}\left|12x+8\right|\ge0\\\left|11y-5\right|\ge0\\\left|13z-y+1\right|\ge0\end{matrix}\right.\Rightarrow\left|12x+8\right|+\left|11y-5\right|+\left|13z-y+1\right|\ge0\)
Mà \(\left|12x+8\right|+\left|11y-5\right|+\left|13z-y+1\right|\le0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|12x+8\right|=0\\\left|11y-5\right|=0\\\left|13z-y+1\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-2}{3}\\y=\dfrac{5}{11}\\z=\dfrac{-6}{11}\end{matrix}\right.\)
Vậy...
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