Để ptr có `2` nghiệm thì `\Delta\ge 0`

`⇔(m+2)^2-4(m+8)\ge 0`

`⇔m^2+4m+4-4m-32\ge 0`

`⇔m^2\ge 28`

`⇒m\ge \sqrt{28}` hoặc `m\le -\sqrt{28}`

Theo Vi-ét: \(\begin{cases} x_1+x_2=m+2(1)\\ x_1x_2=m+8(2) \end{cases}\)

ĐK: `x_1\ge 0;x_2\ge 0`

\(⇔\begin{cases} m+2\ge 0\\ m+8\ge 0\end{cases}⇔m\ge -2⇔m\ge \sqrt{28}\)

Ta có:

\(\begin{array}{l}
x_1\sqrt{x_1}-\sqrt{x_2}=\left(x_1-1\right)\left(x_2-1\right)-7\\
⇔\sqrt{x_1^3}-\sqrt{x_2}=x_1x_2-\left(x_1+x_2\right)-6\\
⇔\sqrt{x_1^3}-\sqrt{x_2}=m+8-m-2-6\\
⇔\sqrt{x_1^3}-\sqrt{x_2}=0\\
⇔\sqrt{x_1^3}=\sqrt{x_2}\\
⇔x_1^3=x_2\left(3\right)\\
(2)(3)⇔x_1^4=m+8\\
⇔x_1=\sqrt[4]{m+8}\\
⇔x_2=\sqrt[4]{\left(m+8\right)^3}
\end{array}\)

Thay vào `(1)`

\(\begin{array}{l}
(1)⇔\sqrt[4]{m+8}+\sqrt[4]{\left(m+8\right)^3}=m+2\\
⇔\sqrt{m+8}+\sqrt{\left(m+8\right)^3}+2\sqrt[4]{\left(m+8\right)^4}=\left(m+2\right)^2\\
⇔\sqrt{m+8}+\sqrt{\left(m+8\right)^3}+2\left(m+8\right)=m^2+4m+4\\
⇔\sqrt{m+8}+\sqrt{\left(m+8\right)^3}=m^2+2m-12\\
⇔m+8+\left(m+8\right)^3+2\sqrt{\left(m+8\right)^4}=\left(m^2+2m-12\right)^2\\
⇔m^3+26m^2+225m+648=m^4+4m^3-20m^2-48m+144\\
⇔\left(m-8\right)\left(m^3+11m^2+42m+63\right)=0\\
\Rightarrow m=8
\end{array}\)

Do `m\ge \sqrt{28}⇒m^3+11m^2+42m+63>0`

Vậy `m=8`