Ta có : \(3x-8y=1\Rightarrow x=\dfrac{8y+1}{3}\)
\(\Rightarrow P=\dfrac{\left(8y+1\right)^2}{9}+y^2=\dfrac{64y^2+16y+1+9y^2}{9}=\dfrac{73y^2+16y+1}{9}\)
\(=\dfrac{73\left(y^2+\dfrac{16}{73}y+\dfrac{1}{73}\right)}{9}=\dfrac{73\left[\left(y^2+\dfrac{16}{73}y+\dfrac{64}{5329}\right)+\dfrac{9}{5329}\right]}{9}\)
\(=\dfrac{73\left[\left(y+\dfrac{8}{73}\right)^2+\dfrac{9}{5329}\right]}{9}\ge\dfrac{73.\dfrac{9}{5329}}{9}=\dfrac{1}{73}\)
Vậy \(MIN_P=\dfrac{1}{73}\) . Dấu \("="\) xảy ra khi \(x=\dfrac{3}{73}\) và \(y=-\dfrac{8}{73}\)
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