Giúp mih vs ạ

Bài 12: Để giải quyết bài toán này, chúng ta sẽ sử dụng phương pháp hoàn chỉnh bình phương để biến đổi các biểu thức thành tổng của các bình phương. Điều này giúp chúng ta dễ dàng tìm ra giá trị của \(x\) và \(y\). Bài 1: \(x^2 + y^2 + 4y + 13 = 6x\) 1. Rearrange the equation: \[ x^2 - 6x + y^2 + 4y + 13 = 0 \] 2. Complete the square for \(x\) and \(y\): \[ (x^2 - 6x + 9) + (y^2 + 4y + 4) = 0 + 9 + 4 - 13 \] \[ (x - 3)^2 + (y + 2)^2 = 0 \] 3. Since the sum of two squares is zero, each square must be zero: \[ (x - 3)^2 = 0 \quad \text{and} \quad (y + 2)^2 = 0 \] \[ x - 3 = 0 \quad \text{and} \quad y + 2 = 0 \] \[ x = 3 \quad \text{and} \quad y = -2 \] Bài 2: \(x^2 + y^2 + 17 = 2x - 8y\) 1. Rearrange the equation: \[ x^2 - 2x + y^2 + 8y + 17 = 0 \] 2. Complete the square for \(x\) and \(y\): \[ (x^2 - 2x + 1) + (y^2 + 8y + 16) = 0 + 1 + 16 - 17 \] \[ (x - 1)^2 + (y + 4)^2 = 0 \] 3. Since the sum of two squares is zero, each square must be zero: \[ (x - 1)^2 = 0 \quad \text{and} \quad (y + 4)^2 = 0 \] \[ x - 1 = 0 \quad \text{and} \quad y + 4 = 0 \] \[ x = 1 \quad \text{and} \quad y = -4 \] Bài 3: \(x^2 + y^2 + 45 = 12y - 6x\) 1. Rearrange the equation: \[ x^2 + 6x + y^2 - 12y + 45 = 0 \] 2. Complete the square for \(x\) and \(y\): \[ (x^2 + 6x + 9) + (y^2 - 12y + 36) = 0 + 9 + 36 - 45 \] \[ (x + 3)^2 + (y - 6)^2 = 0 \] 3. Since the sum of two squares is zero, each square must be zero: \[ (x + 3)^2 = 0 \quad \text{and} \quad (y - 6)^2 = 0 \] \[ x + 3 = 0 \quad \text{and} \quad y - 6 = 0 \] \[ x = -3 \quad \text{and} \quad y = 6 \] Bài 4: \(4x^2 + 9y^2 + 2 = 4x + 6y\) 1. Rearrange the equation: \[ 4x^2 - 4x + 9y^2 - 6y + 2 = 0 \] 2. Complete the square for \(x\) and \(y\): \[ 4(x^2 - x + \frac{1}{4}) + 9(y^2 - \frac{2}{3}y + \frac{1}{9}) = 0 + 1 + 1 - 2 \] \[ 4(x - \frac{1}{2})^2 + 9(y - \frac{1}{3})^2 = 0 \] 3. Since the sum of two squares is zero, each square must be zero: \[ 4(x - \frac{1}{2})^2 = 0 \quad \text{and} \quad 9(y - \frac{1}{3})^2 = 0 \] \[ x - \frac{1}{2} = 0 \quad \text{and} \quad y - \frac{1}{3} = 0 \] \[ x = \frac{1}{2} \quad \text{and} \quad y = \frac{1}{3} \] Bài 5: \(9x^2 + 4y^2 + 26 + 4y = 30x\) 1. Rearrange the equation: \[ 9x^2 - 30x + 4y^2 + 4y + 26 = 0 \] 2. Complete the square for \(x\) and \(y\): \[ 9(x^2 - \frac{10}{3}x + \frac{25}{9}) + 4(y^2 + y + \frac{1}{4}) = 0 + 25 + 1 - 26 \] \[ 9(x - \frac{5}{3})^2 + 4(y + \frac{1}{2})^2 = 0 \] 3. Since the sum of two squares is zero, each square must be zero: \[ 9(x - \frac{5}{3})^2 = 0 \quad \text{and} \quad 4(y + \frac{1}{2})^2 = 0 \] \[ x - \frac{5}{3} = 0 \quad \text{and} \quad y + \frac{1}{2} = 0 \] \[ x = \frac{5}{3} \quad \text{and} \quad y = -\frac{1}{2} \] Bài 6: \(9x^2 + y^2 + 20 = 12x + 8y\) 1. Rearrange the equation: \[ 9x^2 - 12x + y^2 - 8y + 20 = 0 \] 2. Complete the square for \(x\) and \(y\): \[ 9(x^2 - \frac{4}{3}x + \frac{4}{9}) + (y^2 - 8y + 16) = 0 + 4 + 16 - 20 \] \[ 9(x - \frac{2}{3})^2 + (y - 4)^2 = 0 \] 3. Since the sum of two squares is zero, each square must be zero: \[ 9(x - \frac{2}{3})^2 = 0 \quad \text{and} \quad (y - 4)^2 = 0 \] \[ x - \frac{2}{3} = 0 \quad \text{and} \quad y - 4 = 0 \] \[ x = \frac{2}{3} \quad \text{and} \quad y = 4 \] Bài 7: \(x^2 + 49y^2 + 5 + 14y = 4x\) 1. Rearrange the equation: \[ x^2 - 4x + 49y^2 + 14y + 5 = 0 \] 2. Complete the square for \(x\) and \(y\): \[ (x^2 - 4x + 4) + 49(y^2 + \frac{2}{7}y + \frac{1}{49}) = 0 + 4 + 1 - 5 \] \[ (x - 2)^2 + 49(y + \frac{1}{7})^2 = 0 \] 3. Since the sum of two squares is zero, each square must be zero: \[ (x - 2)^2 = 0 \quad \text{and} \quad 49(y + \frac{1}{7})^2 = 0 \] \[ x - 2 = 0 \quad \text{and} \quad y + \frac{1}{7} = 0 \] \[ x = 2 \quad \text{and} \quad y = -\frac{1}{7} \] Bài 8: \(16x^2 + 25y^2 + 13 = 20y + 24x\) 1. Rearrange the equation: \[ 16x^2 - 24x + 25y^2 - 20y + 13 = 0 \] 2. Complete the square for \(x\) and \(y\): \[ 16(x^2 - \frac{3}{2}x + \frac{9}{16}) + 25(y^2 - \frac{4}{5}y + \frac{4}{25}) = 0 + 9 + 4 - 13 \] \[ 16(x - \frac{3}{4})^2 + 25(y - \frac{2}{5})^2 = 0 \] 3. Since the sum of two squares is zero, each square must be zero: \[ 16(x - \frac{3}{4})^2 = 0 \quad \text{and} \quad 25(y - \frac{2}{5})^2 = 0 \] \[ x - \frac{3}{4} = 0 \quad \text{and} \quad y - \frac{2}{5} = 0 \] \[ x = \frac{3}{4} \quad \text{and} \quad y = \frac{2}{5} \] Kết luận: Các giá trị của \(x\) và \(y\) cho từng bài toán lần lượt là: 1. \(x = 3\), \(y = -2\) 2. \(x = 1\), \(y = -4\) 3. \(x = -3\), \(y = 6\) 4. \(x = \frac{1}{2}\), \(y = \frac{1}{3}\) 5. \(x = \frac{5}{3}\), \(y = -\frac{1}{2}\) 6. \(x = \frac{2}{3}\), \(y = 4\) 7. \(x = 2\), \(y = -\frac{1}{7}\) 8. \(x = \frac{3}{4}\), \(y = \frac{2}{5}\)