a)2x^2+xy-y^2-x+2y-1

=2x^2+xy-x-(y-1)^2

=2x^2+x(y-1)-(y-1)^2

=2a^2+ab-b^2         với a=x,b=y-1

=2a^2+2ab-ab-b^2

=(2a-b)(a+b)

=(2x-y+1)(x+y-1)

x²+y²=xy+2 (1) <=>x²+xy+y²=2xy+2 (1')

x³-2x=6y+y³<=>x³-y³=2x+6y<=>(x-y)(x²+xy+y²)=2x+6y (2) 

the (1') vao (2)<=>(x-y)(2xy+2)=2x+6y<=>2x²y-2xy²+2x-2y=2x+6y<=>2x²y-2xy²-8y=0

<=>2y(x²-xy-4)=0 <=>x²-xy-4=0 hoac y=0

truong hop: y=0 thay vao (1) ta dc  x²=2 =>x=   hoac x=

truong hop: x²-xy-4=0

ta dc he moi:

x²+y²=xy+2 (1)

x²   =xy+4    (3)      

lay pt (1)-(3) ta dc y²=   -2 (vo ly)

=>he moi vo nghiem

Vay he pt da cho co 2 nghiem

hằng đẳng thức : x^3 -y^3=(x-y)(x^2-xy+y^2) bạn viết sai ạ

\(a,\hept{\begin{cases}x^2-3y=2\\9y^2-8x=8\end{cases}}\)

\(x^2-3y=2\)

\(y=\frac{1^2-2}{3}\)

\(9-\left(\frac{x^2-2}{3}\right)^2-8x=8\)

\(\Rightarrow x^4-4x^2+4-8x-8=0\)

\(\Rightarrow x^4-4x^2-8x-4=0\)

\(\Rightarrow\left(x^2-2x-2\right)\left(x^2+2x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1+\sqrt{3}\\x=1-\sqrt{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=\frac{2+2\sqrt{3}}{3}\\y=\frac{2-2\sqrt{3}}{3}\end{cases}}\)

Vậy ................................

Nhân 3 lần vế pt 1 lên rồi trừ 2 vê cho nhau

x=7/9 thì y=7/2; x=0 thì y=0