\(Δ'=[-(m+1)]^2-1.m^2=m^2+2m+1-m^2=2m+1\\\text{Pt có 2 nghiệm phân biệt}\,\,→Δ'=2m+1>0\\↔2m>-1\\↔m>-\dfrac{1}{2}\\\text{Theo hệ thức Vi-ét}:\\\begin{cases}x_1+x_2=2(m+1)\\x_1x_2=m^2\end{cases}\\x_1^2+x_2^2=4\sqrt{x_1x_2}\\↔(x_1+x_2)^2-2x_1x_2=4\sqrt{x_1x_2}\\↔[2(m+1)]^2-2m^2=4\sqrt{m^2}\\↔4(m^2+2m+1)-2m^2=4|m|\\↔4m^2+8m+4-2m^2-4|m|=0\\↔2m^2+8m+4-4|m|=0\\↔m^2+4m+2-2|m|=0\\TH1:\,\,m\ge 0→|m|=m\\→m^2+4m+2-2m=0\\↔m^2+2m+1+1=0\\→(m+1)^2+1=0(\text{vô lý})\\→\text{Pt vô nghiệm trong khoảng}\,\,x\ge 0\\TH2:\,\,-\dfrac{1}{2}<m<0→|m|=-m\\→m^2+4m+2+2m=0\\↔m^2+6m+2=0(*)\\Δ'_{(*)}=3^2-1.2.9-2=7>0\\→\text{Pt có 2 nghiệm phân biệt trong khoảng}\,\,-\dfrac{1}{2}<m<0\\→\begin{cases}m_1=-3+\sqrt 7(TM)\\m_2=-3-\sqrt 7(KTM)\end{cases}\\\text{Vậy}\,\,m=-3+\sqrt 7\)
