Jump to content

Photo

$\left\{\begin{matrix} 2x^2+xy-y^2-5x+y+2=0\\x^2+y^2+x+y-4=0 \end{matrix}\right.$

- - - - -

  • Please log in to reply
2 replies to this topic

#1
MazacarJin15

MazacarJin15

    True Blue

  • Thành viên
  • 51 posts
Tìm nghiệm thực của hệ phương trình :
$\left\{\begin{matrix} 2x^2+xy-y^2-5x+y+2=0\\x^2+y^2+x+y-4=0 \end{matrix}\right.$

Edited by MazacarJin15, 23-11-2012 - 23:10.

Posted Image

#2
tieulyly1995

tieulyly1995

    Sĩ quan

  • Thành viên
  • 435 posts

Tìm nghiệm thực của hệ phương trình :
$\left\{\begin{matrix} 2x^2+xy-y^2-5x+y+2=0\\x^2+y^2+x+y-4=0 \end{matrix}\right.$


Nhận thấy :
$PT (1)\Leftrightarrow ( x+y-2)(2x-y-1)=0$

#3
Zaraki

Zaraki

    PQT

  • Phó Quản lý Toán Cao cấp
  • 4273 posts
$$(1) \iff (x+y-2)(2x-y-1)=0 \iff \left [ \begin{array}{l} x+y-2=0 \\ 2x-y-1=0 \end{array} \right.$$
TH1: Với $x+y-2=0 \implies x=2-y$. Thay vào $(2)$ thì $$(2-y)^2+y^2+(2-y)+y-4=0 \iff 2y^2-4y+2=0 \iff (y-1)^2=0 \iff y=1$$
Với $y=1$ thì $x=1$.
TH2: Với $2x-y-1=0 \implies y=2x-1$. Thay vào pt $(2)$ thì $$x^2+(2x-1)^2+x+2x-1-4=0 \iff 5x^2-x-4=0 \iff (5x+4)(x-1)=0 \iff \left [ \begin{array}{l} x= \frac{-4}{5} \\ x=1 \end{array} \right.$$
Với $x= \frac{-4}{5}$ thì $y=- \frac{13}{5}$.
Với $x= 1$ thì $y=1$.

Vậy pt có nghiệm $$\boxed{(x,y)=(1,1), \left( \dfrac{-4}{5}, \dfrac{-13}{5} \right)}$$

Discovery is a child’s privilege. I mean the small child, the child who is not afraid to be wrong, to look silly, to not be serious, and to act differently from everyone else. He is also not afraid that the things he is interested in are in bad taste or turn out to be different from his expectations, from what they should be, or rather he is not afraid of what they actually are. He ignores the silent and flawless consensus that is part of the air we breathe – the consensus of all the people who are, or are reputed to be, reasonable.

 

Grothendieck, Récoltes et Semailles (“Crops and Seeds”). 





1 user(s) are reading this topic

0 members, 1 guests, 0 anonymous users