đặt R3=x(ôm)
\(=>P3=I3^2.x=\dfrac{U^2}{R23^2}.x=\dfrac{12^2x}{\left(8+x\right)^2}=\dfrac{12^2}{\left(\dfrac{8+x}{\sqrt{x}}\right)^2}=\dfrac{12^2}{\left(\dfrac{8}{\sqrt{x}}+\sqrt{x}\right)^2}\)
BDT AM-GM \(=>\left(\dfrac{8}{\sqrt{x}}+\sqrt{x}\right)^2\ge\left(2\sqrt{8}\right)^2=32=>P3\le\dfrac{12^2}{32}=4,5W\)
dấu"=" xảy ra<=>\(x=R3=8\left(om\right)\)