Question

[Quân]

giải hệ phương trình sau: \(\left\{{}\begin{matrix}x^2-4xy+y^2=3\\y^2-3xy=2\end{matrix}\right.\)

Answer 1

[Mysterious Person]

\(\left\{{}\begin{matrix}x^2-4xy+y^2=3\\y^2-3xy=2\end{matrix}\right.\)\(\dfrac{\left(1\right)}{\left(2\right)}\)

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 = \(\dfrac{\dfrac{1}{2}-1}{\sqrt{\dfrac{1}{2}}}\) = - \(\dfrac{\sqrt{2}}{2}\)