$a,$ $(x+6)^4+(x+8)^4=272$

Đặt: $x+7=a$

$⇔(a-1)^4+(a+1)^4=272$

$⇔(a^2-2a+1)^2+(a^2+2a+1)^2=272$

$⇔(a^2+1)^2-4a(a^2+1)+4a^2+(a^2+1)^2+4t(a^2+1)+1=272$

$⇔2(a^2+1)^2+8a^2=272$

$⇔2(a^4+2a^2+1)+8a^2=272$

$⇔2a^4+4a^2+2+8a^2=272$

$⇔(2a^4+4a^2+2+8a^2):2=272:2$

$⇔a^4+2a^2+1+4a^2=136$

$⇔a^4+6a^2-135=0$

$⇔a^4+15a^2-9a^2-135=0$

$⇔a^2(a^2+15)-9(a^2+15)=0$

$⇔(a^2-9)(a^2+15)$

Vì: $a^2+15≥15>0(∀a)$

$⇔a^2-9=0$

$⇔(a-3)(a+3)=0$

$⇔(x+4)(x+10)=0$

$⇔$ \(\left[ \begin{array}{l}x=-4\\x=-10\end{array} \right.\) 

Vậy `S={-4;-10}`

$b,$ $(5-x)^4+(2-x)^4=17$

$⇔(x-5)^4+(x-2)^4=17$

$⇔(x-\dfrac{7}{2}-\dfrac{3}{2})^2+(x-\dfrac{7}{2}+\dfrac{3}{2})^2=17$

Đặt: $x-\dfrac{7}{2}=t$

$⇔(t-\dfrac{3}{2})^4+(t+\dfrac{3}{2})^4=17$

$⇔(t^2-3t+\dfrac{9}{4})^2+(t^2+3t+\dfrac{9}{4})^2=17$

$⇔(t^2+\dfrac{9}{4})^2-6t(t^2+\dfrac{9}{4})+9t^2+(t^2+\dfrac{9}{4})^2+6t(t^2+\dfrac{9}{4})+9t^2=17$

$⇔2(t^2+\dfrac{9}{4})^2+18t^2=17$

$⇔2(t^4+\dfrac{9}{2}t^2+\dfrac{81}{16})^2+18t^2=17$

$⇔2t^4+9t^2+\dfrac{81}{8}+18t^2=17$

$⇔2t^4+27t^2-\dfrac{55}{8}=0$

$⇔t^4+\dfrac{27}{2}.t^2-\dfrac{55}{16}=0$

$⇔(t^2+\dfrac{27}{4})^2-49=0$

$⇔(t^2+\dfrac{27}{4})^2=(±7)^2$

Vì: $t^2+\dfrac{27}{4} \neq -7$

$⇔t^2+\dfrac{27}{4}=7$

$⇔t^2=\dfrac{1}{4}$

$⇔t=±\dfrac{1}{2}$

$⇔$ \(\left[ \begin{array}{l}x=3\\x=4\end{array} \right.\) 

Vậy `S={3;4}`

$a,$ $(x+6)^4+(x+8)^4=272$

Đặt: $x+7=b$

$⇔(b-1)^4+(b+1)^4=272$

$⇔(b^2-2b+1)^2+(b^2+2b+1)^2=272$

$⇔(b^2+1)^2-4b(b^2+1)+4b^2+(a^2+1)^2+4(b^2+1)+1=272$

$⇔2(b^2+1)^2+8b^2=272$

$⇔2(b^4+2b^2+1)+8b^2=272$

$⇔2b^4+4b^2+2+8b^2=272$

$⇔(2b^4+4b^2+2+8b^2):2=272:2$

$⇔b^4+2b^2+1+4b^2=136$

$⇔b^4+6b^2-135=0$

$⇔b^4+15b^2-9b^2-135=0$

$⇔b^2(b^2+15)-9(b^2+15)=0$

$⇔(b^2-9)(b^2+15)$

vì $b^2+15≥15>0(∀a)$

$=>b^2-9=0$

$<=>(b-3)(b+3)=0$

$⇔(x+4)(x+10)=0$

$⇔$ \(\left[ \begin{array}{l}x=-4\\x=-10\end{array} \right.\) 

Vậy `S={-4;-10}`

$b,$ $(5-x)^4+(2-x)^4=17$

$⇔(x-5)^4+(x-2)^4=17$

$⇔(x-\dfrac{7}{2}-\dfrac{3}{2})^2+(x-\dfrac{7}{2}+\dfrac{3}{2})^2=17$

Đặt: $x-\dfrac{7}{2}=a$

$⇔(b-\dfrac{3}{2})^4+(b+\dfrac{3}{2})^4=17$

$⇔(b^2-3t+\dfrac{9}{4})^2+(b^2+3b+\dfrac{9}{4})^2=17$

$⇔(b^2+\dfrac{9}{4})^2-6t(b^2+\dfrac{9}{4})+9b^2+(b^2+\dfrac{9}{4})^2+6b(b^2+\dfrac{9}{4})+9b^2=17$

$⇔2(b^2+\dfrac{9}{4})^2+18b^2=17$

$⇔2(b^4+\dfrac{9}{2}b^2+\dfrac{81}{16})^2+18b^2=17$

$⇔2b^4+9b^2+\dfrac{81}{8}+18b^2=17$

$⇔2b^4+27b^2-\dfrac{55}{8}=0$

$⇔b^4+\dfrac{27}{2}.b^2-\dfrac{55}{16}=0$

$⇔(b^2+\dfrac{27}{4})^2-49=0$

$⇔(b^2+\dfrac{27}{4})^2=(±7)^2$

vì $b^2+\dfrac{27}{4} \neq -7$

$⇔b^2+\dfrac{27}{4}=7$

$⇔b^2=\dfrac{1}{4}$

$⇔b=+\dfrac{1}{2}$ hoặc $-\dfrac{1}{2}$

$⇔$ \(\left[ \begin{array}{l}x=3\\x=4\end{array} \right.\) 

Vậy `S={3;4}`