Chủ đề này có 3 trả lời

[Black Pearl]

Giải HPT 

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

2. $\left\{\begin{matrix} xy-y^2=\sqrt{3y-1}-\sqrt{x+2y-1}\\ x^3y-4xy^2+7xy-5x-y+2=0 \end{matrix}\right.$

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

4. $\left\{\begin{matrix} x^2+(x+y)y+2=9y\\ x+y-7=\frac{y}{x^2+2} \end{matrix}\right.$

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

[TNTFlashNo1]

Giải HPT 

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

2. $\left\{\begin{matrix} xy-y^2=\sqrt{3y-1}-\sqrt{x+2y-1}\\ x^3y-4xy^2+7xy-5x-y+2=0 \end{matrix}\right.$

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

4. $\left\{\begin{matrix} x^2+(x+y)y+2=9y\\ x+y-7=\frac{y}{x^2+2} \end{matrix}\right.$

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

5.Hệ trên $\Leftrightarrow \left\{\begin{matrix} x^{2} =4x-2y& & \\ \left ( 2x-y^{2} \right )^{2} =2y-4& & \end{matrix}\right.$

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

suy ra (x,y)

1.$\sqrt{2x+3y}+\sqrt{2x-3y}=3\sqrt{2y}$

bình phương lên rùi biến đổi ra $y\left ( 5y-2x \right )=0$

thế vào PT 2 suy ra (x,y)

[loolo]

2) Đk: $y\geq \frac{1}{3};x+2y-1\geq 0$

$(1)\Leftrightarrow y(x-y)+\sqrt{x+2y-1}-\sqrt{3y-1}=0$

$\Leftrightarrow (x-y)(y+\frac{1}{\sqrt{x+2y-1}+\sqrt{3y-1}})=0$

$\Leftrightarrow x-y=0$ (trong ngoặc luôn dương với y>$\frac{1}{3}$)

$\Leftrightarrow x=y$

...................

[basketball123]

Giải HPT 

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

2. $\left\{\begin{matrix} xy-y^2=\sqrt{3y-1}-\sqrt{x+2y-1}\\ x^3y-4xy^2+7xy-5x-y+2=0 \end{matrix}\right.$

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

4. $\left\{\begin{matrix} x^2+(x+y)y+2=9y\\ x+y-7=\frac{y}{x^2+2} \end{matrix}\right.$

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

3) ĐK $x\geq 1;y\geq 1$

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

Thay vào $(2)\Leftrightarrow \sqrt{2y}+\sqrt{y-1}=3\Leftrightarrow (\sqrt{2y}-2)+(\sqrt{y-1}-1)=0\Leftrightarrow \frac{2(y-2)}{\sqrt{2y}+2}+\frac{y-2}{\sqrt{y-1}+1}=0\Leftrightarrow (y-2)(\frac{2}{\sqrt{2y}+2}+\frac{1}{\sqrt{y-1}+1})=0\Rightarrow y=2\Rightarrow x=5$

Vậy $(x;y)=(5;2)$