Question

\(|12x+8|+|11y-5|+|13z-y+1|\le0\)

Answer 1

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 \(\left\{{}\begin{matrix}\left|12x+8\right|=0\\\left|11y-5\right|=0\\\left|13z-y+1\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}12x+8=0\\11y-5=0\\13z-y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{8}{12}=-\dfrac{2}{3}\\y=\dfrac{5}{11}\\z=\dfrac{-1+y}{13}=\dfrac{-1+\dfrac{5}{11}}{13}=-\dfrac{6}{143}\end{matrix}\right.\)

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Answer 2

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 \(\left|12x+8\right|+\left|11y-5\right|+\left|13z-y+1\right|\le0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|12x+8\right|=0\\\left|11y-5\right|=0\\\left|13z-y+1\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-2}{3}\\y=\dfrac{5}{11}\\z=\dfrac{-6}{11}\end{matrix}\right.\)

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