a: \(\Leftrightarrow\left(2m-4\right)^2-4\left(m^2-3\right)>=0\)

\(\Leftrightarrow4m^2-16m+16-4m^2+12>=0\)

=>-16m>=-28

hay m<=7/4

b: \(\Leftrightarrow16m^2-4\left(2m-1\right)\left(2m+3\right)=0\)

\(\Leftrightarrow16m^2-4\left(4m^2+4m-3\right)=0\)

=>4m-3=0

hay m=3/4

c: \(\Leftrightarrow\left(4m-2\right)^2-4\cdot4\cdot m^2< 0\)

=>-16m+4<0

hay m>1/4

\(\Leftrightarrow\left[\left(x-2\right)^2-4\right]^2-3\left(x-2\right)^2+m=0\)

\(\left(x-2\right)^2=t\ge0\Rightarrow pt\Leftrightarrow\left(t-4\right)^2-3t+m=0\)

\(\Leftrightarrow t^2-11t+16+m=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta=11^2-4\left(16+m\right)>0\\x_1+x_2=11>0\left(tm\right)\\x_1x_2=16+m>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< \dfrac{57}{4}\\m< 16\end{matrix}\right.\Leftrightarrow m< \dfrac{57}{4}\)

b) Thay x=2 vào pt, ta được:

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

\(\Leftrightarrow4m^2-4-4m+m^2+m+4=0\)

\(\Leftrightarrow5m^2-3m=0\)

\(\Leftrightarrow m\left(5m-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=0\\m=\dfrac{3}{5}\end{matrix}\right.\)

Áp dụng hệ thức Vi-et, ta được:

\(x_1+x_2=\dfrac{2m}{m^2-1}\)

\(\Leftrightarrow\left[{}\begin{matrix}x_2+2=0\\x_2+2=\dfrac{6}{5}:\left(\dfrac{36}{25}-1\right)=\dfrac{30}{11}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x_2=-2\\x_2=\dfrac{8}{11}\end{matrix}\right.\)

\(\Delta=\left(2m-2\right)^2-4\cdot2\cdot\left(m+2-\sqrt{2}\right)\)

\(=4m^2-8m+4-8m-8+8\sqrt{2}\)

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

Để phương trình có nghiệm kép thì \(4m^2-16m+8\sqrt{2}-4=0\)

=>\(m^2-4m+2\sqrt{2}-1=0\)

=>\(\Delta=\left(-4\right)^2-4\left(2\sqrt{2}-1\right)=16-8\sqrt{2}+4=20-8\sqrt{2}>0\)

=>Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}m=\dfrac{4-\sqrt{20-8\sqrt{2}}}{2}=2-\sqrt{5-2\sqrt{2}}\\m=2+\sqrt{5-2\sqrt{2}}\end{matrix}\right.\)

1:

\(=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{2}{3\sqrt{x}-6}\right):\dfrac{2\sqrt{x}+3}{3\sqrt{x}}\)

\(=\dfrac{3+2\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{3\sqrt{x}}{2\sqrt{x}+3}=\dfrac{1}{\sqrt{x}-2}\)

\(\Leftrightarrow\sqrt{2x^2-2\left(m+4\right)x+5m+10}=x-3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3\ge0\\2x^2-2\left(m+4\right)x+5m+10=x^2-6x+9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge3\\x^2-2\left(m+1\right)x+5m+1=0\left(1\right)\end{matrix}\right.\)

Pt đã cho có nghiệm khi (1) có ít nhất 1 nghiệm thỏa mãn \(x\ge3\)

- Để (1) có nghiệm \(\Leftrightarrow\Delta'=\left(m+1\right)^2-\left(5m+1\right)\ge0\Leftrightarrow m^2-3m\ge0\Rightarrow\left[{}\begin{matrix}m\ge3\\m\le0\end{matrix}\right.\) (1)

- Để 2 nghiệm của (1) thỏa mãn \(x_1\le x_2< 3\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x_1-3\right)\left(x_2-3\right)>0\\\dfrac{x_1+x_2}{2}< 3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x_1x_2-3\left(x_1+x_2\right)+9>0\\x_1+x_2< 6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5m+1-6\left(m+1\right)+9>0\\2\left(m+1\right)< 6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< 4\\m< 2\end{matrix}\right.\) \(\Rightarrow m< 2\)

\(\Rightarrow\) Để pt có ít nhất 1 nghiệm thỏa mãn \(x\ge3\) thì \(m\ge2\) (2)

Kết hợp (1); (2) \(\Rightarrow m\ge3\)

ĐK; \(-1\le x\le3\)

Đặt \(\sqrt{-x^2+2x+3}=t\left(0\le t\le2\right)\)

\(pt\Leftrightarrow m+1=-x^2+2x+3+4\sqrt{-x^2+2x+3}\)

\(\Leftrightarrow m+1=f\left(t\right)=t^2+4t\)

\(f\left(0\right)=0;f\left(2\right)=12\)

Yêu cầu bài toán thỏa mãn khi \(minf\left(t\right)\le m+1\le maxf\left(t\right)\)

\(\Leftrightarrow0\le m+1\le12\)

\(\Leftrightarrow-1\le m\le11\)