Có $VT=\left| {2x-3} \right|+\left| {7-2x} \right|\geq \left| {2x-3+7-2x} \right|=4$$VP=\sqrt{-x^{2}+4x-4+16}=\sqrt{-(x-2)^{2}+16}\leq 4$Do đó:$VT=VP=4\Rightarrow x=2$
ĐK : $12+4x-x^{2} \geq 0 <=> -2 \leq x \leq 6$Có $VT=\left| {2x-3} \right|+\left| {7-2x} \right|\geq \left| {2x-3+7-2x} \right|=4$$VP=\sqrt{-x^{2}+4x-4+16}=\sqrt{-(x-2)^{2}+16}\leq 4$Do đó:$VT=VP=4\Rightarrow x=2$
(tm)