Vì: \(\left\{{}\begin{matrix}\left|3x-4\right|\ge0\forall x\\\left|5y+5\right|\ge0\forall y\end{matrix}\right.\)
=> Để |3x - 4| + |5y + 5| = 0 thì:
\(\left\{{}\begin{matrix}\left|3x-4\right|=0\\\left|5y+5\right|=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}3x-4=0\\5y+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=-1\end{matrix}\right.\)
Vây..........................................
|3x - 4| + |5y + 5| = 0
Ta có: \(\left\{{}\begin{matrix}\left|3x-4\right|\ge0\forall x\in R\\\left|5y+5\right|\ge0\forall x\in R\end{matrix}\right.\)
Mà |3x - 4| + |5y + 5| = 0 (Theo bài cho)
\(\Rightarrow\left\{{}\begin{matrix}\left|3x-4\right|=0\\\left|5y+5\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-4=0\\5y+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=0+4\\5y=0-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=4\\5y=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{-5}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\x=-1\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{4}{3}\\x=-1\end{matrix}\right.\)