Toán Lớp 8: Tìm giá trị nhỏ nhất
$x^{2}$+4$y^{2}$-2xy+2x+10y+23
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Toán Lớp 8: Tìm giá trị nhỏ nhất
$x^{2}$+4$y^{2}$-2xy+2x+10y+23
Comments ( 1 )
\quad x^2 + 4y^2 – 2xy + 2x + 10y + 23\\
= (x^2 – 2xy + y^2 + 2x – 2y + 1) + (3y^2 + 12y + 12) + 10\\
= (x -y + 1)^2 + 3(y + 2)^2 + 10\\
\text{Ta có:}\\
\quad \begin{cases}(x-y+1)^2\geqslant 0\quad \forall x,y\\(y+2)^2 \geqslant \quad \forall y\end{cases}\\
\Leftrightarrow (x -y + 1)^2 + 3(y + 2)^2 + 10\geqslant 10\\
\text{Dấu = xảy ra}\ \Leftrightarrow \begin{cases}x – y + 1 =0\\y + 2 = 0\end{cases}\Leftrightarrow \begin{cases}x = -3\\y = -2\end{cases}\\
\text{Vậy}\ GTNN\ (x^2 + 4y^2 – 2xy + 2x + 10y + 23) = 10\Leftrightarrow (x;y)=(-3;-2)
\end{array}\)