Question

Giải phương trình nghiệm nguyên

a) 3x^2−4y^2=18

b) 19x^2+28y^2=2001

c) x^2=2y^2−8y+3

d) x^2+y^2-4x+4y=1

Answer 1

a. 3x² - 4y² = 18

<=> \(\left\{{}\begin{matrix}3x^2=18+4y^2\\4y^2=-\left(3x^2-18\right)\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=\sqrt{\dfrac{18+4y^2}{3}}\\y=\sqrt{\dfrac{-3x^2+18}{4}}\end{matrix}\right.\)

b, c, d tương tự nhé

Answer 2

b. 19x² + 28y² = 2001

<=> \(\left\{{}\begin{matrix}19x^2=2001-28y^2\\28y^2=2001-19x^2\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=\sqrt{\dfrac{2001-28y^2}{19}}\\y=\sqrt{\dfrac{2001-19x^2}{28}}\end{matrix}\right.\)

c. x² = 2y² - 8y + 3

<=> \(\left\{{}\begin{matrix}x=\sqrt{2y^2-8y+3}\\8y=2y^2+3-x^2\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=\sqrt{2y^2-8y+3}\\y=\dfrac{2y^2+3-x^2}{8}\end{matrix}\right.\)

d. x² + y² - 4x + 4y = 1

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

<=> \(\left\{{}\begin{matrix}x=\sqrt{1-y^2+4x-4y}\\y=\sqrt{1-x^2+4x-4y}\end{matrix}\right.\)