Question

[Thi Bùi]

x^2-2xy+6y^2-8x-12y+36=0 tìm x, y

Answer 1

[l҉o҉n҉g҉ d҉z҉]

x² - 2xy + 6y² - 8x - 12y + 36 = 0

⇔ ( x² - 2xy + y² - 8x + 8y + 16 ) + ( 5y² - 20y + 20 ) = 0

⇔ [ ( x² - 2xy + y² ) - ( 8x - 8y ) + 16 ] + 5( y² - 4y + 4 ) = 0

⇔ [ ( x - y )² - 2.( x - y ).4 + 4² ] + 5( y - 2 )² = 0

⇔ ( x - y - 4 )² + 5( y - 2 )² = 0

Ta có : \(\hept{\begin{cases}\left(x-y-4\right)^2\ge0\forall x,y\\5\left(y-2\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x-y-4\right)^2+5\left(y-2\right)^2\ge0\forall x,y\)

Dấu "=" xảy ra ⇔ \(\hept{\begin{cases}x-y-4=0\\y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\y=2\end{cases}}\)

Vậy x = 6 ; y = 2