Giải hệ phương trình
$\begin{cases}(1-y)\sqrt{x-y}+x=2 + (x-y-1)\sqrt{y} \\ 2y^{2}-3x+6y+1 = 2\sqrt{x-2y}-\sqrt{4x-5y-3} \end{cases}$
bài này có đáp án r mà bn:)) –  Bloody's Rose 09-06-16 03:45 PM
Điều kiện {y0x2y4x5y+3()
Ta có:
(1)(yx+1)(xy1)+(xy1)(1y)=0
(1y)(xy1)(1xy+1+1y+1)=0
[y=1y=x1
*** Với y=1 phương trình (2) trở thành 93x=0x=3
*** Với y=x1 điều kiện (*) trở thành 1x2 Phương trình 2 trở thành:
2x2x3=2x
2(x2x1)+(x12x)=0
(x2x1)(2+1x1+2x)=0
x2x1=0x=1±52
Đối chiếu (*) và kết hợp trường hợp trên ta có nghiệm của hệ là 

Bạn cần đăng nhập để có thể gửi đáp án

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