Đáp án + Giải thích các bước giải:
`2019|2x-1|+5(x+2y)^{24}=0`
Vì $\left\{\begin{matrix}|2x-1|≥0& \\(x+2y)^{24}≥0& \end{matrix}\right.$
`→` $\left\{\begin{matrix}2019|2x-1|≥0& \\5(x+2y)^{24}≥0& \end{matrix}\right.$
`→2019|2x-1|+5(x+2y)^{24}≥0`
Mà theo đề bài : `2019|2x-1|+5(x+2y)^{24}=0`
`→` $\left\{\begin{matrix}2019|2x-1|=0& \\5(x+2y)^{24}=0& \end{matrix}\right.$
`→` $\left\{\begin{matrix}|2x-1|=0& \\(x+2y)^{24}=0& \end{matrix}\right.$
`→` $\left\{\begin{matrix}2x-1=0& \\x+2y=0& \end{matrix}\right.$
`→` $\left\{\begin{matrix}x=\frac{1}{2}& \\2y=-x& \end{matrix}\right.$
`→` $\left\{\begin{matrix}x=\frac{1}{2}& \\2y=-\frac{1}{2}& \end{matrix}\right.$
`→` $\left\{\begin{matrix}x=\frac{1}{2}& \\y=-\frac{1}{4}& \end{matrix}\right.$
Vậy `(x;y)=((1)/(2);-(1)/(4))`
