Question

\(\left|3x+2y\right|+\left|4y-1\right|\le0\)

Answer 1

Ta có: \(\left|3x+2y\right|\ge0\) và \(\left|4y-1\right|\ge0\)

Nên: \(\left|3x+2y\right|+\left|4y-1\right|\le0\) khi:

\(\left\{{}\begin{matrix}3x+2y=0\\4y-1=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3x+2y=0\\4y=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3x+2y=0\\y=\dfrac{1}{4}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3x+2\cdot\dfrac{1}{4}=0\\y=\dfrac{1}{4}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{6}\\y=\dfrac{1}{4}\end{matrix}\right.\)

Vậy (x;y) thỏa mãn là: \(\left(-\dfrac{1}{6};\dfrac{1}{4}\right)\)

Answer 2