`(4-x)^5+(x-2)^5=32`
`⇔-(x-4)^5+(x-2)^5=32`
Đặt `x-3=y`
`⇔-(y-1)^5+(y+1)^5=32`
`⇔-(y^5−5y^4+10y^3−10y^2+5y−1)+(y^5+5y^4+10y^3+10y^2+5y+1)=32`
`⇔-y^5+5y^4-10y^3+10y^2-5y+1+y^5+5y^4+10y^3+10y^2+5y+1=32`
`⇔10y^4+20y^2+2=32`
`⇔10y^4+20y^2+2-32=0`
`⇔10y^4+20y^2-30=0`
`⇔10(y^4+2y^2-3)=0`
`⇔y^4+2y^2-3=0`
`⇔y^4+y^3-y^3-y^2+3y^2-3=0`
`⇔ y^3(y+1)-y^2(y+1)+3(y^2-1)=0`
`⇔ y^3(y+1)-x^2(y+1)+3(y-1)(y+1)=0`
`⇔(y+1)(y^3-y^2+3y-3)=0`
`⇔(y+1)[y^2(y-1)+3(y-1)]=0`
`⇔(y+1)(y-1)(y^2+3)=0`
`⇔(y+1)(y-1)=0` (vì `y^2+3\ge3>0AAx`)
`⇒`\(\left[ \begin{array}{l}y+1=0\\y-1=0\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}y=-1\\y=1\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x-3=-1\\x-3=1\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x=2\\x=4\end{array} \right.\)
Vậy tập nghiệm của phương trình `S={2;4}.`
