Quay lại đáp án

$A=(x^2+xy+\frac{y^2}{4}+\frac{59}{4}y^2+8x+y+1992$ $=(x+\frac{y}{2})^2+\frac{59}{4}y^2+8x+y+1992$ $=(x+\frac{y}{2}+4)^2+\frac{59}{4}y^2-3y+1992-16$ $=(x+\frac{y}{2}+4)^2+\frac{59}{4}(y^2-\frac{12}{59}y)+1976$ $=(x+\frac{y}{2}+4)^2+\frac{59}{4}(y-\frac{6}{59})^2+1976-\frac{9}{59}$$\Rightarrow minA=1975\tfrac{50}{59}\Leftrightarrow y=\frac{6}{59};x=-4\tfrac{3}{59}$

$A=(x^2+xy+\frac{y^2}{4}+\frac{59}{4}y^2+8x+y+1992$ $=(x+\frac{y}{2})^2+\frac{59}{4}y^2+8x+y+1992$ $=(x+\frac{y}{2}+4)^2+\frac{59}{4}y^2-3y+1992-16$ $=(x+\frac{y}{2}+4)^2+\frac{59}{4}(y^2-\frac{12}{59}y)+1976$ $=(x+\frac{y}{2}+4)^2+\frac{59}{4}(y-\frac{6}{59})^2+1976-\frac{9}{59}$$\Rightarrow \min A=1975\tfrac{50}{59}\Leftrightarrow y=\frac{6}{59};x=-4\tfrac{3}{59}$