Ta có

\(\left\{{}\begin{matrix}\sqrt{2x^2-4x+3}=\sqrt{2\left(x-1\right)^2+1}\ge\sqrt{1}=1\\\sqrt{3x^2-6x+7}=\sqrt{3\left(x-1\right)^2+4}\ge\sqrt{4}=2\end{matrix}\right.\) \(\forall x\)

\(\Rightarrow VT=\sqrt{2x^2-4x+3}+\sqrt{3x^2-6x+7}\ge3\) \(\forall x\)

Lại có \(VP=2-x^2+2x=3-\left(x-1\right)^2\le3\) \(\forall x\)

\(\Rightarrow\sqrt{2x^2-4x+3}+\sqrt{3x^2-6x+7}=2-x^2+2x\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2\left(x-1\right)^2+1}=1\\\sqrt{3\left(x-1\right)^2+4}=2\\3-\left(x-1\right)^2=3\end{matrix}\right.\)

\(\Leftrightarrow\left(x-1\right)^2=0\Rightarrow x=1\)

Vậy pt có nghiệm duy nhất \(x=1\)

\(ĐK:x\in R\)

Đặt \(x^2-2x=a\), PTTT:

\(-a+\sqrt{6a+7}=0\\ \Leftrightarrow\sqrt{6a+7}=a\\ \Leftrightarrow a^2-6a-7=0\\ \Leftrightarrow\left[{}\begin{matrix}a=7\\a=-1\left(loại.do.a=\sqrt{6a+7}\ge0\right)\end{matrix}\right.\\ \Leftrightarrow a=7\\ \Leftrightarrow x^2-2x-7=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1+2\sqrt{2}\\x=1-2\sqrt{2}\end{matrix}\right.\)

=>|x-1|+|x-3|=1

TH1: x<1

Pt sẽ la 1-x+3-x=1

=>4-2x=1

=>x=3/2(loại)

TH2: 1<=x<3

Pt sẽ là x-1+3-x=1

=>2=1(loại)

TH3: x>=3

Pt sẽ là x-1+x-3=1

=>2x-4=1

=>2x=5

=>x=5/2(loại)

Đặt \(\sqrt{-x^2+2x+15}=t\Rightarrow0\le t\le4\)

BPT trở thành:

\(-4t\ge-t^2+2+m\)

\(\Leftrightarrow t^2-4t-2\ge m\)

\(\Rightarrow m\le\min\limits_{\left[0;4\right]}\left(t^2-4t-2\right)\)

Xét \(f\left(t\right)=t^2-4t-2\) trên \(\left[0;4\right]\)

\(-\dfrac{b}{2a}=2\in\left[0;4\right]\)

\(f\left(0\right)=f\left(4\right)=-2\) ; \(f\left(2\right)=-6\)

\(\Rightarrow f\left(t\right)_{min}=-6\Rightarrow m\le-6\)

A