8 replies to this topic

[Mirror282]

1)$\left\{\begin{matrix} xy+x+y=x^{2}-2y^{2} & & \\ x\sqrt{2y}-y\sqrt{x-1}=2x-2y & & \end{matrix}\right.$

 

2)$\left\{\begin{matrix} x^{4}+2x^{3}y+2x^{2}y^{2}=2x+9 & & \\ x^{2}+2xy=6x+6 & & \end{matrix}\right.$

 

3)$\left\{\begin{matrix} x^{2}+y^{2}+xy+1=4y & & \\ (x^{2}+1)(x+y-2)=y & & \end{matrix}\right.$

 

4)$\left\{\begin{matrix} xy+2=3x^{3} & & \\2xy^{3}-3x=-y^{2} & & \end{matrix}\right.$

 

5)$\left\{\begin{matrix} 1+xy+\sqrt{xy}=x & & \\ \frac{1}{x\sqrt{x}}+y\sqrt{y}=\frac{1}{\sqrt{x}}+3\sqrt{y} & & \end{matrix}\right.$

 

6)$\left\{\begin{matrix} x^{4}+y^{4}=x^{3}y+xy^{3}& & \\ \sqrt{4-x}+\sqrt{y+1}-\sqrt{3x-x^{2}+4}=1 & & \end{matrix}\right.$

 

7)$\left\{\begin{matrix} x^{2}+2x+y^{2}+y=3-xy & & \\ xy+x+2y=1 & & \end{matrix}\right.$

 

8)$\left\{\begin{matrix} x^{3}-6x^{2}y+9xy^{2}-4y^{3}=0 & & \\ \sqrt{x-y}+\sqrt{x+y}=2 & & \end{matrix}\right.$

 

9)$\left\{\begin{matrix} x+3y+3\sqrt{xy+y^{2}}=xy+y^{2}+y\sqrt{x+y} & & \\x+y=xy+x^{2} & & \end{matrix}\right.$

 

10)$\left\{\begin{matrix} x(x+1)+\frac{1+y}{y^{2}}=4 & & \\ (xy+1) (x^{2}y^{2}+1)-4y^{3}=0 & & \end{matrix}\right.$

 

11)$\left\{\begin{matrix} x^{4}+x^{2}y^{2}-y^{2}=y^{3}+x^{2}y+x^{2} & & \\2y^{3}-\sqrt{5-2x^{2}}-1=0 & & \end{matrix}\right.$

 

Edited by Mirror282, 23-02-2015 - 02:52.

[GeminiKid]

6)$\left\{\begin{matrix} x^{4}+y^{4}=x^{3}y+xy^{3}& & \\ \sqrt{4-x}+\sqrt{y+1}-\sqrt{3x-x^{2}+4}=1 & & \end{matrix}\right.$

$PT(1)\Leftrightarrow \left ( x-y \right )^{2}\left ( x^{2} +xy+y^{2}\right )\Leftrightarrow x=y$ 

Thế vào PT(2) có $PT(2)\Leftrightarrow \sqrt{4-x}+\sqrt{x+1}-\sqrt{\left ( 4-x \right )\left ( x+1 \right )}= 1$ 

Đặt $\sqrt{4-x}= a\geq 0,\sqrt{x+1}= b\geq 0$

$\Rightarrow \left\{\begin{matrix} a+b-ab= 1 & & \\ a^{2} +b^{2}=5& & \end{matrix}\right.$ 

giải hệ này chắc dễ rồi

[GeminiKid]

 

4)$\left\{\begin{matrix} xy+2=3x^{3} & & \\2xy^{3}-3x=-y^{2} & & \end{matrix}\right.$

 

 

Hệ đã cho$\Leftrightarrow \left\{\begin{matrix} x^{2}\left ( 3x-y^{2} \right )+x^{2}y^{2}-xy-2 & & \\ 3x-y^{2}= 2xy^{3}& & \end{matrix}\right.$

thế PT dưới vào PT trên đc PT 1 ẩn xy=> dễ rồi

Edited by GeminiKid, 23-02-2015 - 14:05.

[GeminiKid]

 

 

8)$\left\{\begin{matrix} x^{3}-6x^{2}y+9xy^{2}-4y^{3}=0 & & \\ \sqrt{x-y}+\sqrt{x+y}=2 & & \end{matrix}\right.$

 

 

 

 

$PT(1)\Leftrightarrow \left ( x-y \right )^{2}\left ( x-4y \right )= 0$

[kudoshinichihv99]

 

3)$\left\{\begin{matrix} x^{2}+y^{2}+xy+1=4y & & \\ (x^{2}+1)(x+y-2)=y & & \end{matrix}\right.$

 

 

Chia cả 2 pt cho y

Đặt $\ \frac{x^{2}+1}{y}= a$

       x+y=b

=>hệ

[kudoshinichihv99]

 

 

2)$\left\{\begin{matrix} x^{4}+2x^{3}y+2x^{2}y^{2}=2x+9 & & \\ x^{2}+2xy=6x+6 & & \end{matrix}\right.$

 

 

pt1 <=>x⁴+2xy(x²+2xy)=2x+9

pt2=>x²+2xy=6x+6 và 2xy=6x+6-x²  thay vào 1 rồi giải tiếp

[GeminiKid]

1)$\left\{\begin{matrix} xy+x+y=x^{2}-2y^{2} & & \\ x\sqrt{2y}-y\sqrt{x-1}=2x-2y & & \end{matrix}\right.$

 

$PT(1)\Leftrightarrow \left ( x+y \right )\left ( x-2y-1 \right )$

[KantouA11]

$(7)$ Cộng 2 phương trình lại, ta được:

$x^2 + 2xy + y ^2 + 3x + 3y = 4$

$\Leftrightarrow (x+y)^2 + 3(x+y) -4=0$

$\Leftrightarrow x+ y = 1 \vee x+y = -4$

 

$(9)$ ĐK: $\left\{\begin{matrix} y(x+y) \geq 0\\ (x+y)\geq 0 \end{matrix}\right. \Rightarrow \left\{\begin{matrix} y\geq 0\\ x+y \geq 0 \end{matrix}\right.$

$(2) \Leftrightarrow (1-x)(x+y)=0 \Leftrightarrow x=1 \vee x+y=0$

Edited by KantouA11, 23-02-2015 - 18:46.

[GeminiKid]

 

11)$\left\{\begin{matrix} x^{4}+x^{2}y^{2}-y^{2}=y^{3}+x^{2}y+x^{2} & & \\2y^{3}-\sqrt{5-2x^{2}}-1=0 & & \end{matrix}\right.$

 

 

 

 

 

$PT(1)\Leftrightarrow \left ( x^{2}+y^{2} \right )\left ( x^{2} -y-1\right )= 0$