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Bài giải
a, \(\left|x+3\right|+\left|y-1\right|=0\)
Mà \(\hept{\begin{cases}\left|x+3\right|\ge0\forall x\\\left|y-1\right|\ge0\forall x\end{cases}}\Rightarrow\hept{\begin{cases}\left|x+3\right|=0\\\left|y-1\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)
Vậy \(\left(x\text{ ; }y\right)=\left(-3\text{ ; }1\right)\)
b, \(\left|x+5\right|+\left|y+1\right|\le0\)
Mà \(\hept{\begin{cases}\left|x+5\right|\ge0\forall x\\\left|y+1\right|\ge0\end{cases}}\Rightarrow\text{ }\left|x+5\right|+\left|y+1\right|=0\)
Dấu " = " xảy ra khi \(\hept{\begin{cases}\left|x+5\right|=0\\\left|y+1\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-5\\y=-1\end{cases}}\)
Vậy \(\left(x\text{ ; }y\right)=\left(-5\text{ ; }-1\right)\)
#)Giải :
\(\left|2-x\right|+2=x\)
\(\Rightarrow\orbr{\begin{cases}\left|2-x\right|=x\\2=x\end{cases}\Rightarrow x=2}\)
Vậy \(x=2\)
\(\left|x-1\right|\left|-x-1\right|=0\)
\(\Rightarrow\orbr{\begin{cases}\left|x-1\right|=0\\\left|-x-1\right|=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}}\)
Vậy \(x\in\left\{1;-1\right\}\)
a: \(\Leftrightarrow\left\{{}\begin{matrix}3x-2>-4\\3x-2< 4\end{matrix}\right.\Leftrightarrow-\dfrac{2}{3}< x< 2\)
c: \(\Leftrightarrow\left[{}\begin{matrix}3x-1>5\\3x-1< -5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -\dfrac{4}{3}\end{matrix}\right.\)
d: \(\Leftrightarrow\left[{}\begin{matrix}3x+1>x-2\\3x+1< -x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x>-3\\4x< 1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-\dfrac{3}{2}\\x< \dfrac{1}{4}\end{matrix}\right.\)
\(\left|x-7\right|+\left|3-x\right|=\dfrac{12}{\left|y+1\right|+3}\)
Ta có: \(\left\{{}\begin{matrix}\left|x-7\right|+\left|3-x\right|\ge\left|x-7+3-x\right|=4\\\dfrac{12}{\left|y+1\right|+3}\le\dfrac{12}{3}=4\end{matrix}\right.\)
Mà theo đề bài: \(\left|x-7\right|+\left|3-x\right|=\dfrac{12}{\left|y+1\right|+3}\)
\(\Rightarrow\left|x-7\right|+\left|3-x\right|=\dfrac{12}{\left|y+1\right|+3}=4\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}3\le x\le7\\y=-1\end{matrix}\right.\)