\(\frac{a^2+b^2}{a-2b}=2\Rightarrow a^2+b^2-2a+4b=0\Rightarrow\left(a-1\right)^2+\left(b+2\right)^2=5\)
Đặt \(a-1=x,b+2=y\Rightarrow x^2+y^2=5\), khi đó:
\(P=8a+4b=8\left(x+1\right)+4\left(y-2\right)=8x+4y\)
Áp dụng BĐT Cauchy-schwarz, ta có:
\(P^2=\left(8x+4y\right)^2\le\left(8^2+4^2\right)\left(x^2+y^2\right)=400\)
\(\Rightarrow P\le20\)
Vậy \(MaxP=20\) khi ...