\(x^2-2\left(\right.m+1\left.\right)x+m^2+4=0\)

\(\Delta = \left[\right. - 2 \left(\right. m + 1 \left.\right) \left]\right.^{2} - 4 \left(\right. m^{2} + 4 \left.\right) = 8 m - 12 > 0\)

\(m > \frac{3}{2}\)

\(x_{1}^{2} = 2 \left(\right. m + 1 \left.\right) x_{1} - m^{2} - 4\)

\(x_{1}^{2} + 2 \left(\right. m + 1 \left.\right) x_{2} = 3 m^{2} + 16\)

\(\Leftrightarrow 2 \left(\right. m + 1 \left.\right) x_{1} - m^{2} - 4 + 2 \left(\right. m + 1 \left.\right) x_{2} = 3 m^{2} + 16\) \(\Leftrightarrow 2 \left(\right. m + 1 \left.\right) \left(\right. x_{1} + x_{2} \left.\right) - m^{2} - 4 = 3 m^{2} + 16\)

\(x_{1} + x_{2} = 2 \left(\right. m + 1 \left.\right)\)

\(\Rightarrow 4 \left(\right. m + 1 \left.\right)^{2} - m^{2} - 4 = 3 m^{2} + 16\)

\(\Leftrightarrow 4 m^{2} + 8 m - m^{2} = 3 m^{2} + 16\)

\(\Leftrightarrow 8 m = 16\)

\(\Leftrightarrow m = 2\)

Vì \(2>\frac{3}{2}\) nên \(m=2\) thỏa mãn đề bài.

Vậy \(m=2\)

\(x^{3} + y^{3} + z^{3} - 3 x y z = \frac{1}{2} \left(\right. x + y + z \left.\right) \left[\right. \left(\right. x - y \left.\right)^{2} + \left(\right. y - z \left.\right)^{2} + \left(\right. z - x \left.\right)^{2} \left]\right. .\)

Do

\(x^{3} + y^{3} + z^{3} = 3 x y z\)

nên

\(\frac{1}{2} \left(\right. x + y + z \left.\right) \left[\right. \left(\right. x - y \left.\right)^{2} + \left(\right. y - z \left.\right)^{2} + \left(\right. z - x \left.\right)^{2} \left]\right. = 0.\)

Vì \(x , y , z > 0\) nên \(x + y + z > 0\).

⇒

Do đó

Vậy nghiệm là

Từ \(\left(\right. x + y \left.\right)^{2} = x^{2} + y^{2} + 2 x y\)

⇒ \(100 = 58 + 2 x y\)

\(x y = 21.\)

Vậy \(x , y\) là nghiệm của:

\(t^{2} - 10 t + 21 = 0\) \(\left(\right. t - 3 \left.\right) \left(\right. t - 7 \left.\right) = 0.\)

Do đó: \(\left(\right.x,y\left.\right)=\left(\right.3,7\left.\right)\&\left(\right.7,3\left.\right).\)

hi