Cho phương trình: x²+2(m+2)x+2m+3=0

a) Khi m = 1, phương trình trở thành: x² + 2(1+2)x + 2(1) + 3 = 0
Simplifying the equation, we get: x² + 6x + 5 = 0
To solve this quadratic equation, we can factor it as: (x + 1)(x + 5) = 0
Setting each factor equal to zero, we have: x + 1 = 0 or x + 5 = 0
Solving these equations, we find: x = -1 or x = -5
So the solutions to the equation when m = 1 are x = -1 and x = -5.

b) To prove that the equation has solutions for all values of m, we need to show that the discriminant (Δ) of the quadratic equation is always greater than or equal to zero.

The discriminant (Δ) is given by the formula: Δ = b² - 4ac, where a, b, and c are the coefficients of the quadratic equation.

In this case, a = 1, b = 2(m+2), and c = 2m+3.

Substituting these values into the discriminant formula, we have: Δ = [2(m+2)]² - 4(1)(2m+3)
Expanding and simplifying, we get: Δ = 4(m² + 4m + 4) - 8m - 12
Simplifying further, we have: Δ = 4m² + 16m + 16 - 8m - 12
Combining like terms, we get: Δ = 4m² + 8m + 4

To prove that the equation has solutions for all values of m, we need to show that Δ ≥ 0 for all values of m.

Let's consider the discriminant Δ as a quadratic function of m: f(m) = 4m² + 8m + 4.

To find the minimum value of this quadratic function, we can use the vertex formula: m = -b/2a.

In this case, a = 4 and b = 8, so the vertex formula gives us: m = -8/(2*4) = -1.

Since the coefficient of m² is positive, the parabola opens upwards and the vertex represents the minimum point of the quadratic function.

Substituting m = -1 into the quadratic function, we have: f(-1) = 4(-1)² + 8(-1) + 4 = 4 - 8 + 4 = 0.

Since the minimum value of the quadratic function is 0, and the coefficient of m² is positive, the quadratic function is always greater than or equal to 0 for all values of m.

Therefore, the discriminant Δ is always greater than or equal to 0, which means the quadratic equation x² + 2(m+2)x + 2m+3 = 0 has solutions for all values of m.