Câu(2) Cho phương trinh x^2+2(m+1)x+m^2=0 . Tìm m có 2 nghiệm phân biệt trong đó có 1 nghiệm bằng -2

Let the given quadratic equation be<br />$$x^2 + 2(m+1)x + m^2 = 0$$<br />Let $x_1$ and $x_2$ be the roots of the equation.<br />We are given that one root is -2. Let $x_1 = -2$.<br /><br />Substituting $x = -2$ into the equation, we get:<br />$$(-2)^2 + 2(m+1)(-2) + m^2 = 0$$<br />$$4 - 4(m+1) + m^2 = 0$$<br />$$4 - 4m - 4 + m^2 = 0$$<br />$$m^2 - 4m = 0$$<br />$$m(m - 4) = 0$$<br />Thus, $m = 0$ or $m = 4$.<br /><br />If $m=0$, the equation becomes $x^2 + 2x = 0$, which factors as $x(x+2) = 0$. The roots are $x=0$ and $x=-2$. These are distinct roots.<br /><br />If $m=4$, the equation becomes $x^2 + 10x + 16 = 0$. This factors as $(x+2)(x+8) = 0$. The roots are $x=-2$ and $x=-8$. These are distinct roots.<br /><br />Therefore, the values of $m$ for which the equation has two distinct roots, one of which is -2, are $m=0$ and $m=4$.<br /><br />Final Answer: The final answer is $\boxed{0,4}$