Đáp án:

 {(x+2y+1)(x+2y+2)=0xy+y2+3y+1=0​

TH1:

x+2y+1=0  ⟹  x=−2y−1x + 2y + 1 = 0 \implies x = -2y - 1x+2y+1=0⟹x=−2y−1

Thay vào:

(−2y−1)y+y2+3y+1=0(-2y - 1)y + y^2 + 3y + 1 = 0 (−2y−1)y+y2+3y+1=0 −2y2−y+y2+3y+1=0-2y^2 - y + y^2 + 3y + 1 = 0 −2y2−y+y2+3y+1=0 −y2+2y+1=0-y^2 + 2y + 1 = 0 −y2+2y+1=0 y2−2y−1=0y^2 - 2y - 1 = 0 y2−2y−1=0 Δ=4+4=8,Δ=22\Delta = 4 + 4 = 8, \quad \sqrt{\Delta} = 2\sqrt{2}Δ=4+4=8,Δ​=22​ y1=1+2,y2=1−2y_1 = 1 + \sqrt{2}, \quad y_2 = 1 - \sqrt{2}y1​=1+2​,y2​=1−2​ x1=−2(1+2)−1=−3−22x_1 = -2(1 + \sqrt{2}) - 1 = -3 - 2\sqrt{2}x1​=−2(1+2​)−1=−3−22​ x2=−2(1−2)−1=−3+22x_2 = -2(1 - \sqrt{2}) - 1 = -3 + 2\sqrt{2}x2​=−2(1−2​)−1=−3+22​

TH2:

x+2y+2=0  ⟹  x=−2y−2x + 2y + 2 = 0 \implies x = -2y - 2x+2y+2=0⟹x=−2y−2

Thay vào:

(−2y−2)y+y2+3y+1=0(-2y - 2)y + y^2 + 3y + 1 = 0 (−2y−2)y+y2+3y+1=0 −2y2−2y+y2+3y+1=0-2y^2 - 2y + y^2 + 3y + 1 = 0 −2y2−2y+y2+3y+1=0 −y2+y+1=0-y^2 + y + 1 = 0 −y2+y+1=0 y2−y−1=0y^2 - y - 1 = 0 y2−y−1=0 Δ=1+4=5,Δ=5\Delta = 1 + 4 = 5, \quad \sqrt{\Delta} = \sqrt{5}Δ=1+4=5,Δ​=5​ y1=1+52,y2=1−52y_1 = \frac{1 + \sqrt{5}}{2}, \quad y_2 = \frac{1 - \sqrt{5}}{2}y1​=21+5​​,y2​=21−5​​ x1=−2⋅1+52−2=−3−5x_1 = -2 \cdot \frac{1 + \sqrt{5}}{2} - 2 = -3 - \sqrt{5}x1​=−2⋅21+5​​−2=−3−5​ x2=−2⋅1−52−2=−3+5x_2 = -2 \cdot \frac{1 - \sqrt{5}}{2} - 2 = -3 + \sqrt{5}x2​=−2⋅21−5​​−2=−3+5​

Kết luận:

{(−3−22,  1+2)(−3+22,  1−2)(−3−5,  1+52)(−3+5,  1−52)\begin{cases} (-3 - 2\sqrt{2},\; 1 + \sqrt{2}) \\ (-3 + 2\sqrt{2},\; 1 - \sqrt{2}) \\ (-3 - \sqrt{5},\; \frac{1 + \sqrt{5}}{2}) \\ (-3 + \sqrt{5},\; \frac{1 - \sqrt{5}}{2}) \end{cases}⎩⎨⎧​(−3−22​,1+2​)(−3+22​,1−2​)(−3−5​,21+5​​)(−3+5​,21−5​​)​

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