Tìm các số nguyên để biểu thức cho giá trị nguyên

Trong bài viết này, chúng ta sẽ tìm các số nguyên để biểu thức \(A = 3 + \frac{1}{x+3}\) có giá trị nguyên. Điều này đòi hỏi chúng ta phải tìm xem có những giá trị nào của x mà khi thay vào biểu thức, ta thu được một số nguyên. Để giải quyết vấn đề này, chúng ta sẽ xem xét từng giá trị của x một cách cẩn thận. Đầu tiên, hãy xem xét giá trị x = -2. Thay x = -2 vào biểu thức, ta có: \(A = 3 + \frac{1}{-2+3} = 3 + \frac{1}{1} = 3 + 1 = 4\) Như vậy, khi x = -2, giá trị của biểu thức A là 4. Điều này không phải là một số nguyên. Tiếp theo, hãy xem xét giá trị x = 2. Thay x = 2 vào biểu thức, ta có: \(A = 3 + \frac{1}{2+3} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + \frac{1}{5} = 3 + 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