Vì \(\hept{\begin{cases}\left(4x-3\right)^2\ge0\\\left|5y+7,5\right|\ge0\end{cases}\Rightarrow}\left(4x-3\right)^2+\left|5y+7,5\right|\ge0\)
\(\Rightarrow\left(4x-3\right)^2+\left|5y+7,5\right|+17,5\ge17,5\)
Dấu "=" xảy ra khi \(\left(4x-3\right)^2=\left|5y+7,5\right|=0\)
(4x-3)2=0 <=> 4x-3=0 <=> 4x=3 <=> x=3/4|5y+7,5|=0 <=> 5y+7,5=0 <=> 5y=-7,5 <=> y=-3/2Vậy ......