1.\(m=2\)\(\Leftrightarrow x^2-2\left(2+1\right)x+3.2-3=0\)\(\Leftrightarrow x^2-6x+3=0\)\(\Delta=\left(-6\right)^2-4.3=36-12=24>0\)=> pt có 2 nghiệm\(\left\{{}\begin{matrix}x_1=\dfrac{6+2\sqrt{6}}{2}=3+\sqrt{6}\\x_2=\dfrac{6-2\sqrt{6}}{2}=3-\sqrt{6}\end{matrix}\right.\)b.Theo hệ thức Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1.x_2=3m-3\end{matrix}\right.\)\(\sqrt{x_1-1}+\sqrt{x_2-1}=4\)\(\Leftrightarrow x_1-1+x_2-1+2\sqrt{\left(x_1-1\right)\left(x_2-1\right)}=16\)\(\Leftrightarrow x_1+x_2+2\sqrt{x_1x_2-x_1-x_2+1}=18\)\(\Leftrightarrow2m+2+2\sqrt{3m-3-\left(2m+2\right)+1}=18\)\(\Leftrightarrow2m+2\sqrt{m+4}=16\)\(\Leftrightarrow m+\sqrt{m+4}=8\)\(\Leftrightarrow\sqrt{m+4}=8-m\)\(\Leftrightarrow m+4=64-16m+m^2\)\(\Leftrightarrow m^2-17m+60=0\)\(\Delta=\left(-17\right)^2-4.60=49>0\)\(\rightarrow\left\{{}\begin{matrix}x_1=\dfrac{17+\sqrt{49}}{2}=12\\x_2=\dfrac{17-\sqrt{49}}{2}=5\end{matrix}\right.\)Vậy \(m=12;5\)Câu trả lời: 1.\(m=2\)\(\Leftrightarrow x^2-2\left(2+1\right)x+3.2-3=0\)\(\Leftrightarrow x^2-6x+3=0\)\(\Delta=\left(-6\right)^2-4.3=36-12=24>0\)=> pt có 2 nghiệm\(\left\{{}\begin{matrix}x_1=\dfrac{6+2\sqrt{6}}{2}=3+\sqrt{6}\\x_2=\dfrac{6-2\sqrt{6}}{2}=3-\sqrt{6}\end{matrix}\right.\)b.Theo hệ thức Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1.x_2=3m-3\end{matrix}\right.\)\(\sqrt{x_1-1}+\sqrt{x_2-1}=4\)\(\Leftrightarrow x_1-1+x_2-1+2\sqrt{\left(x_1-1\right)\left(x_2-1\right)}=16\)\(\Leftrightarrow x_1+x_2+2\sqrt{x_1x_2-x_1-x_2+1}=18\)\(\Leftrightarrow2m+2+2\sqrt{3m-3-\left(2m+2\right)+1}=18\)\(\Leftrightarrow2m+2\sqrt{m+4}=16\)\(\Leftrightarrow m+\sqrt{m+4}=8\)\(\Leftrightarrow\sqrt{m+4}=8-m\)\(\Leftrightarrow m+4=64-16m+m^2\)\(\Leftrightarrow m^2-17m+60=0\)\(\Delta=\left(-17\right)^2-4.60=49>0\)\(\rightarrow\left\{{}\begin{matrix}x_1=\dfrac{17+\sqrt{49}}{2}=12\\x_2=\dfrac{17-\sqrt{49}}{2}=5\end{matrix}\right.\)Vậy \(m=12;5\)