a/ ĐK: $1\le x\le 5$
Xét VT:
$\sqrt{3x^2-18x+28}\\=\sqrt{3x^2-18x+27+1}\\=\sqrt{3(x^2-6x+9)+1}\\=\sqrt{3(x-3)^2+1}(1)$
Ta có: $3(x-3)^2\ge 0$
$↔3(x-3)^2+1\ge 1\\↔\sqrt{3(x-3)^2+1}\ge 1$
$\sqrt{4x^2-24x+45}\\=\sqrt{4x^2-24x+36+9}\\=\sqrt{(2x-6)^2+9}(2)$
Ta có: $(2x-6)^2\ge 0$
$↔(2x-6)^2+9\ge 9\\↔\sqrt{(2x-6)^2+9}\ge 3$
(1)+(2) $→\sqrt{3x^2-18x+28}+\sqrt{4x^2-24x+45}\ge 4(\star)$
Xét VP:
$-x^2+6x-5\\=-x^2+6x-9+4\\=-(x-3)^2+4$
Ta có: $-(x-3)^2\le 0$
$↔-(x-3)^2+4\le 4\\↔-x^2+6x-5\le 4(\star\star)$
$(\star)(\star\star)→$ Dấu "=" xảy ra khi $\begin{cases}x-3=0\\2x-6=0\\x-3=0\end{cases}$
$↔x=3(TM)$
Vậy phương trình có tập nghiệm $S=\{3\}$
b/ ĐK: $-1-\sqrt 6\le x\le -1+\sqrt 6$
Xét VT:
$\sqrt{3x^2+6x+7}\\=\sqrt{3x^2+6x+3+4}\\=\sqrt{3(x^2+2x+1)+4}\\=\sqrt{3(x+1)^2+4}$
Ta có: $3(x+1)^2\ge 0$
$↔3(x+1)^2+4\ge 4\\↔\sqrt{3x^2+6x+7}\ge 2(1)$
$\sqrt{5x^2+10x+21}\\=\sqrt{5x^2+10x+5+16}\\=\sqrt{5(x^2+2x+1)+16}\\=\sqrt{5(x+1)^2+16}$
Ta có: $5(x+1)^2\ge0$
$↔5(x+1)^2+16\ge 16\\↔\sqrt{5x^2+10x+21}\ge 4(2)$
(1)+(2) $→\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}\ge 6(\star)$
Xét VP:
$5-2x-x^2\\=-x^2-2x-1+6\\=-(x+1)^2+6$
Ta có: $-(x+1)^2\le 0$
$↔-(x+1)^2+6\le 6(\star\star)$
$(\star)(\star\star)→$ Dấu "=" xảy ra khi $x+1=0$
$↔x=-1(TM)$
Vậy phương trình có tập nghiệm $S=\{-1\}$
