Giải thích các bước giải:
`(3/4-x)^2020+(3/7.y-5)^4<=0`
Có $\left\{\begin{matrix}(\dfrac34-x)^{2020}\ge0\\(\dfrac37.y-5)^4\ge0\end{matrix}\right.$
`=>(3/4-x)^2020+(3/7.y-5)^4>=0`
Mà `(3/4-x)^2020+(3/7.y-5)^4<=0`
`=>(3/4-x)^2020+(3/7.y-5)^4=0`
`=>`$\left\{\begin{matrix}\dfrac34-x=0\\\dfrac37.y-5=0\end{matrix}\right.$`=>`$\left\{\begin{matrix}x=\dfrac34\\\dfrac37.y=5\end{matrix}\right.$`=>`$\left\{\begin{matrix}x=\dfrac34\\y=\dfrac{35}3\end{matrix}\right.$
Vậy `x=3/4;y=35/3.`
