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Ta có: \(\left|x-3\right|^{2014}\ge0;\left|6+2y\right|^{2015}\ge0\)
\(\Rightarrow\left|x-3\right|^{2014}+\left|6+2y\right|^{2015}\ge0\)
Mà theo đề: \(\left|x-3\right|^{2014}+\left|6+2y\right|^{2015}\le0\)
=> \(\left|x-3\right|^{2014}+\left|6+2y\right|^{2015}=0\)
=> \(\left|x-3\right|=\left|6+2y\right|=0\)
=> \(x-3=6+2y=0\)
=> \(x=3;y=-3\).
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\(\Leftrightarrow-\dfrac{3}{4}< =x< =\dfrac{1}{2}\)
hay x=0
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Ta có:
\(\left|x-13\right|\ge0;\left|2y-8\right|\ge0\) ( định nghĩa trị tuyệt đối luôn không âm )
\(\Rightarrow\left|x-13\right|+\left|2y-8\right|\ge0\)
Mà \(\left|x-13\right|+\left|2y-8\right|\le0\) nên \(\left|x-13\right|+\left|2y-8\right|=0\)
Khi đó:\(x-13=0;2y-8=0\)
\(\Rightarrow x=13;y=4\)
Vậy x=13;y=4
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Vì \(\left(\frac{1}{2}x-5\right)^{10}\ge0\)và \(\left(y^2-\frac{1}{4}\right)^{20}\ge0\)
nên \(\left(\frac{1}{2}x-5\right)^{10}+\left(y^2-\frac{1}{4}\right)^{20}=0\)
<=>\(\hept{\begin{cases}\frac{1}{2}x-5=0\\y^2-\frac{1}{4}=0\end{cases}}\)<=>\(\hept{\begin{cases}x=10\\y=\pm\frac{1}{2}\end{cases}}\)
Ta có:\(\hept{\begin{cases}\left\{\frac{1}{2}x-5\right\}^{10}\ge0\forall x\\\left\{y^2-\frac{1}{4}\right\}^{20}\ge0\forall y\end{cases}}\)
Mà \(\left\{\frac{1}{2}x-5\right\}^{10}+\left\{y^2-\frac{1}{4}\right\}^{20}\le0\)
\(\Rightarrow\left\{\frac{1}{2}x-5\right\}^{10}+\left\{y^2-\frac{1}{4}\right\}^{20}=0\)
\(\Leftrightarrow\hept{\begin{cases}\left\{\frac{1}{2}x-5\right\}^{10}=0\\\left\{y^2-\frac{1}{4}\right\}^{20}=0\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{2}x-5=0\\y^2-\frac{1}{4}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{1}{2}x=5\\y^2=\frac{1}{4}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=10\\y=\pm\frac{1}{2}\end{cases}}}\)
Vậy \(x=10;y=\pm\frac{1}{2}\)
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\(\left|5x-2\right|\le0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-2\le0\\5x-2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{2}{5}\\x\ge-\dfrac{2}{5}\end{matrix}\right.\)
\(\text{Vì: }\)\(x\in Z\)
\(S=\left\{0\right\}\)
Vì
\(\left|3x+9\right|\ge0\)
\(\left(2y+4\right)^2\ge0\)
\(\Rightarrow\left|3x+9\right|+\left(2y+4\right)\ge0\)
Để \(\left|3x+9\right|+\left(2y+4\right)^2\le0\Rightarrow\orbr{\begin{cases}\left|3x+9\right|=0\\\left(2y+4\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\y=-2\end{cases}}}\)
Vậy x = - 3; y = - 2
Ta có:
\(\left|3x+9\right|>\)hoặc bằng 0.\(\left(1\right)\)
\(\left(2y+4\right)^2>=0\)\(\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\left|3x+9\right|=0\); \(\left(2y+4\right)^2=0\)
\(\left|3x+9\right|=0\)
\(\Rightarrow3x+9=0\)
\(\Rightarrow3x=-9\)
\(\Rightarrow x=-3\)
ta lại có:
\(\left(2y+4\right)^2=0\)
\(\Rightarrow2y+4=0\)
\(\Rightarrow2y=-4\)
\(\Rightarrow y=-2\)
Vậy \(x=-3;y=-2\).