\begin{cases}x^{4}+x^{3}y + x^{2}y^{2}=1 \\ x^{3}y-x^{2} + xy=-1 \end{cases}

Giúp mình nha các bạn!

pt1 <=> x³(x - y) = (1 - xy)(1 + xy)
pt2 <=> (1 + xy)/(1 - xy) = x²
từ 2 pt trên dễ dàng => (1 - xy)² = x(x - y) (*)
tiếp tục từ pt 1 =>(x² - 1)(x² + 1) = x²y(x - y) = xy*x(x - y) (**)
từ pt 2 => xy = (x² - 1)/(x² + 1) (***)
thay (***) và (*) vào (**) ta đc
(x² - 1)(x² + 1) = [(x² - 1)*(1 - xy)²]/(x² + 1)
<=> x² = 1 => ............
hoặc
(x² + 1)² = ( 1 - xy)² <=> (x² - xy)(x² + 1 + 1 - xy) = 0 <=> (x² - xy) = 0
( cái kia loại vì x² - xy + 2 = (1 - xy)² + 2 > 0 )
với x² = 1 => y = ???
x² - xy => x = y ( x = 0 VN)
bạn có thể tham khảo thêm ở các link sau:
http://diendantoanhoc.net/topic/69337-leftbeginarraylx4-x3yx2y21-x3y-x2xy-1endarrayright/
http://diendantoanhoc.net/topic/87191-leftbeginmatrix-x4-x3yx2y21-x3y-x2xy-1-endmatrixright/
http://k2pi.net.vn/tags.php?tag=gia%CC%89i+h%C3%AA%CC%A3+x4-x3y+x2y2%3D1+va%CC%80+x3y-x2+xy%3D-1
ẹc ck là thánh soi òi –  Đức Anh 08-05-16 05:15 PM

Bạn cần đăng nhập để có thể gửi đáp án

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