Question

Answer 1

Ta có: \(M=x^2-2x\left(y+1\right)+3y^2+2025\)

\(=x^2-2x\left(y+1\right)+\left(y+1\right)^2+3y^2-\left(y+1\right)^2+2025\)

\(=\left(x-y-1\right)^2+3y^2-y^2-2y-1+2025\)

\(=\left(x-y-1\right)^2+2y^2-2y+2024\)

\(=\left(x-y-1\right)^2+2\left(y^2-y+\frac14\right)-\frac12+2024\)

\(=\left(x-y-1\right)^2+2\left(y-\frac12\right)^2+2023,5\ge2023,5\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y-\frac12=0\\ x-y-1=0\end{cases}\Rightarrow\begin{cases}y=\frac12\\ x=y+1=\frac12+1=\frac32\end{cases}\)