\(\left\{{}\begin{matrix}x^2-4xy+y^2=3\\y^2-3xy=2\end{matrix}\right.\)\(\dfrac{\left(1\right)}{\left(2\right)}\)
ta lấy (1) - (2) ta có : -xy + x2 = 1 \(\Leftrightarrow\) -xy = 1-x2 \(\Leftrightarrow\) xy = x2-1 (3)
\(\Rightarrow\) y = \(\dfrac{x^2-1}{x}\) (4)
thay (3) và (4) vào (2) ta có :
\(\left(\dfrac{x^2-1}{x}\right)\)2 -3(x2-1) = 2
\(\Leftrightarrow\) \(\dfrac{\left(x^2-1\right)^2}{x^2}\) - \(\dfrac{3\left(x^2-1\right)}{1}\) = 2
\(\Leftrightarrow\) \(\dfrac{\left(x^2-1\right)^2+x^2\left(-3x^2+3\right)}{x^2}\) = 2
\(\Leftrightarrow\) \(\dfrac{x^4-2x^2+1-3x^4+3x^2}{x^2}\) = 2
\(\Leftrightarrow\) \(\dfrac{-2x^4+x^2+1}{x^2}\) = 2
\(\Leftrightarrow\) -2x4+x2+1 = 2x2
\(\Leftrightarrow\) -2x4-x2+1 = 0
đặc x2 = t (t\(\ge\) 0 )
ta có : a-b+c = -2+1+1= 0
\(\Rightarrow\) phương trình có 2 nghiệm phân biệt
t1= -1 (loại) ; t2 = -\(\dfrac{c}{a}\) = \(\dfrac{-1}{-2}\) = \(\dfrac{1}{2}\) (tmđk)
vậy t = x2 = \(\dfrac{1}{2}\) \(\Leftrightarrow\) x = \(\pm\) \(\sqrt{\dfrac{1}{2}}\)
x = - \(\sqrt{\dfrac{1}{2}}\) \(\Rightarrow\) y = \(\dfrac{\dfrac{1}{2}-1}{-\sqrt{\dfrac{1}{2}}}\) = \(\dfrac{\sqrt{2}}{2}\)
x = \(\sqrt{\dfrac{1}{2}}\) \(\Rightarrow\) y = \(\dfrac{\dfrac{1}{2}-1}{\sqrt{\dfrac{1}{2}}}\) = - \(\dfrac{\sqrt{2}}{2}\)
