Question

Cho \(A=\left(x-2y\right)^2+\left(x-3\right)^2+\left(y-1\right)^2+3\). The least possible value of A is...

Answer 1

Lời giải:

Ta có:

\(A=(x-2y)^2+(x-3)^2+(y-1)^2+3\)

\(=x^2+4y^2-4xy+x^2-6x+9+y^2-2y+1+3\)

\(=2x^2+5y^2-4xy-6x-2y+13\)

\(=2(x^2-2xy+y^2)-6x-2y+3y^2+13\)

\(=2(x-y)^2-2.3(x-y)-8y+3y^2+13\)

\(=2[(x-y)^2-3(x-y)+\frac{3^2}{2^2}]+3(y^2-\frac{8}{3}y+\frac{4^2}{3^2})+\frac{19}{6}\)

\(=2(x-y-\frac{3}{2})^2+3(y-\frac{4}{3})^2+\frac{19}{6}\)

\(\geq 0+0+\frac{19}{6}=\frac{19}{6}\)

Vậy GTNN của $A$ là \(\frac{19}{6}\)

Dấu "=" xảy ra khi \(\left\{\begin{matrix} x-y-\frac{3}{2}=0\\ y-\frac{4}{3}=0\end{matrix}\right.\Leftrightarrow x=\frac{17}{6}; y=\frac{4}{3}\)