CHÚC BẠN HỌC TỐT

\(3x^2+2y^2=5xy\)

\(\Leftrightarrow3x^2+2y^2-5xy=0\)

\(\Leftrightarrow2\left(x^2-2xy+y^2\right)+x^2-xy=0\)

\(\Leftrightarrow2\left(x-y\right)^2+x\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left[2\left(x-y\right)+x\right]=0\)

\(\Leftrightarrow\left(x-y\right)\left(3x-2y\right)=0\)

\(\Leftrightarrow3x-2y=0\Leftrightarrow x=\dfrac{2y}{3}\) Thay vào S

\(\Rightarrow S=\dfrac{y+\dfrac{4y}{3}}{y-\dfrac{4y}{3}}=-7\)

\(x\ge xy+1\Rightarrow1\ge y+\dfrac{1}{x}\ge2\sqrt{\dfrac{y}{x}}\Rightarrow\dfrac{y}{x}\le\dfrac{1}{4}\)

\(Q^2=\dfrac{x^2+2xy+y^2}{3x^2-xy+y^2}=\dfrac{\left(\dfrac{y}{x}\right)^2+2\left(\dfrac{y}{x}\right)+1}{\left(\dfrac{y}{x}\right)^2-\dfrac{y}{x}+3}\)

Đặt \(\dfrac{y}{x}=t\le\dfrac{1}{4}\) 

\(Q^2=\dfrac{t^2+2t+1}{t^2-t+3}=\dfrac{t^2+2t+1}{t^2-t+3}-\dfrac{5}{9}+\dfrac{5}{9}\)

\(Q^2=\dfrac{\left(4t-1\right)\left(t+6\right)}{9\left(t^2-t+3\right)}+\dfrac{5}{9}\le\dfrac{5}{9}\)

\(\Rightarrow Q_{max}=\dfrac{\sqrt{5}}{3}\) khi \(t=\dfrac{1}{4}\) hay \(\left(x;y\right)=\left(2;\dfrac{1}{2}\right)\)

\(A=2\left(x^2+y^2\right)+\left(8y^2+\dfrac{1}{2}z^2\right)+\left(8x^2+\dfrac{1}{2}z^2\right)\ge2.2\sqrt{x^2y^2}+2\sqrt{8x^2.\dfrac{1}{2}z^2}+2.\sqrt{8x^2.\dfrac{1}{2}z^2}=4\left(xy+yz+zx\right)=4\)

\(A_{min}=4\) khi \(\left(x;y;z\right)=\left(\dfrac{1}{3};\dfrac{1}{3};\dfrac{4}{3}\right)\)