Toán Lớp 8: Giải phương trình nghiệm nguyên
x²+2003x+2004y+y²=xy+2004xy²+2005
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Toán Lớp 8: Giải phương trình nghiệm nguyên
x²+2003x+2004y+y²=xy+2004xy²+2005
Comments ( 1 )
$\min K = 1975 \Leftrightarrow (x;y) = \left(5;\dfrac{7}{3}\right)$
Lời giải và giải thích chi tiết:
$\begin{array}{l}K = 2x^2 + 9y^2- 6xy – 6x – 12y + 2004\\ = \dfrac{1}{2}(4x^2 + 9y^2 +9 – 12xy + 18y – 12x)+ \dfrac{9}{2}\left(y^2 – \dfrac{14}{3}y + \dfrac{49}{9}\right) + 1975\\ = \dfrac{1}{2}(2x – 3y – 3)^2 + \dfrac{9}{2}\left(y – \dfrac{7}{3}\right)^2 + 1975\\ Ta\,\,có:\\ \begin{cases}(2x – 3y – 3)^2 \geq 0, \,\forall x,y\\\left(y – \dfrac{7}{3}\right)^2 \geq 0, \,\forall y\end{cases}\\ Do\,\,đó:\\ \dfrac{1}{2}(2x – 3y – 3)^2 + \dfrac{9}{2}\left(y – \dfrac{7}{3}\right)^2 + 1975 \geq 1975\\ \text{Dấu = xảy ra}\, \Leftrightarrow \begin{cases}2x – 3y – 3 = 0\\y – \dfrac{7}{3} = 0\end{cases} \Leftrightarrow \begin{cases}x = 5\\y = \dfrac{7}{3}\end{cases}\\ Vậy\,\,\min K = 1975 \Leftrightarrow (x;y) = \left(5;\dfrac{7}{3}\right) \end{array}$