\(C=x^4-8xy-x^3y+x^2y^2-xy^3+y^4+2022\)

\(=\left(x^4-x^3y-xy^3+y^4\right)+\left(x^2y^2-8xy+16\right)+2006\)

\(=\left(x-y\right)^2\left(x^2+xy+y^2\right)+\left(xy-4\right)^2+2006\ge2006\forall\left(x,y\right)\)

Dấu "=" xảy ra khi 

\(\left\{{}\begin{matrix}x-y=0\\xy-4=0\end{matrix}\right.\Leftrightarrow x=y=\pm2}\)

Bài 1:

\(A=\left(x-y\right)^2+\left(x+1\right)^2+\left(y-5\right)^2+2011\)

\(=2x^2+2y^2-2xy+2x-10y+2037\)

\(=2\left(x^2+\left(\dfrac{-y}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-xy+x-\dfrac{y}{2}\right)+\dfrac{3y^2}{2}-9y-\dfrac{1}{2}+2037\)

\(=2\left(x-\dfrac{y}{2}+\dfrac{1}{2}\right)^2+\dfrac{3}{2}\left(y-3\right)^2-\dfrac{27}{2}-\dfrac{1}{2}+2037\ge2023\forall\left(x,y\right)\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{y}{2}+\dfrac{1}{2}=0\\y-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)

\(x^2+1=2x+\sqrt{3x-1}\) (không cần điệu kiện)

\(\Leftrightarrow x^2-\left(3x-1\right)=-x+\sqrt{3x-1}\)

\(\Leftrightarrow x^2-\left(\sqrt{3x-1}\right)^2=-\left(x-\sqrt{3x-1}\right)\)

\(\Leftrightarrow\left(x-\sqrt{3x-1}\right)\left(x+\sqrt{3x-1}\right)=-\left(x-\sqrt{3x-1}\right)\)

\(\Leftrightarrow\left(x-\sqrt{3x-1}\right)\left(x+1+\sqrt{3x-1}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{3x-1}=0\left(1\right)\\x+1+\sqrt{3x-1}=0\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x=\sqrt{3x-1}\Leftrightarrow\left\{{}\begin{matrix}x^2=3x-1\\x\ge0\end{matrix}\right.\Leftrightarrow x=\dfrac{3\pm\sqrt{5}}{2}\)

\(\left(2\right)\Leftrightarrow-x-1=\sqrt{3x-1}\Leftrightarrow\left\{{}\begin{matrix}\left(-x-1\right)^2=3x-1\\-x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-x+2=0\\x\le-1\end{matrix}\right.\)(vô nghiệm)

Vậy phương trình có hai nghiệm \(x_{1,2}=\dfrac{3\pm\sqrt{5}}{2}\)