1, Tìm x, y biết:
a, 2x2 - 8x + y2 + 2y + 9 = 0
b, x2 + y2 + x - xy + 1/2 = 0
c, x2 - 4xy + 92 + 9 = 2xy + 6x - x2
d, 5x2 + 5y2 + 8xy + 2y - 2x + 2 = 0
e, 4x2 + 13y2 +12xy + 4x - 2y + 5 =0
f, x2 + 2y2 - 2xy + 2x + 2 - 4y = 0
g, 8x3 + (x + 8)2 = 8(x + 2)(x2 - 2x + 4)
h, (3x +1)(9x2 + 1 - 3x) - (3 - x)2 = (3x - 2)3
i, 2x2 + 5y2 - 4xy - 4x - 14y + 29 = 0
j, 4x2 + 9y2 - 12x - 32y - 2xy + 44 = 0
f) x2 + 2y2 - 2xy + 2x + 2 - 4y =0
<=>x2 + y2 - 2xy+2x-2y+y2-2y+1+1=0
<=>(x-y)2+2(x-y)+1+(y-1)2=0
<=>(x-y+1)2+(y-1)2=0
<=>y=1;x=0
Bạn học thầy Trung phải k nè~~~~
Busted :))))
a)\(2x^2-8x+y^2+2y+9=0\)
\(\Leftrightarrow\left(2x^2-8x+8\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow2\left(x-2\right)^2+\left(y+1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-2=0\\y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=-1\end{cases}}\)
Vậy \(x=2\)và \(y=-1\)
