Question

Помогите решить уравнение

Answer 1

$(x+5)^{4}+(x+3)^{4}=2\\t=x+4=>x=t-4\\x+5=t-4+5=t+1\\x+3=t-4+3=t-1\\(t+1)^{4}+(t-1)^{4}=2\\(t^{2}+2t+1)^{2}+(t^{2}-2t+1)^{2}=2\\t^{4}+4t^{3}+4t^{2}+2t^{2}+4t+1+t^{4}-4t^{3}+4t^{2}+2t^{2}-4t+1-2=0\\2t^{4}+12t^{2}=0\\t^{4}+6t^{2}=0\\t^{2}(t^{2}+6)=0\\t^{2}=0=>t=0\\t^{2}+6=0=>t^{2}=-6 -neud.\\t=0=>x+4=0=>x=-4\\Otvet:x=-4.$