(x-2)^4 + (x-3)^4 = 1

[latex](x-2)^{4} + (x-3)^{4} =1 \ (x-2)^{4} + (x-3)^{4} = 1^{4} \ sqrt{(x-2)^{4}} + sqrt{(x-3)^{4}} = sqrt{{1^{4}}} \(x-2)^{2} + (x-3)^{2} = 1^{2} \ x^{2} -4x+4+ x^{2} -6x+9=1 \ 2 x^{2} -10x+13-1=0 \ 2 x^{2} -10x+12=0 \ D= -10^{2} -4*2*12=100-96=4= 2^{2} \ x_{1} = frac{10+2}{2*2} =3; \ x_{2} = frac{10-2}{2*2} =2. \ [/latex]Ответ: х₁ = 3; х₂ = 2.

Замена (х-3) = t[latex](t+1)^4+t^4=1\ t^4+4t^3+6t^2+4t+1+t^4=1\ 2t^4+4t^3+6t^2+4t=0\ 2t(t^3+2t^2+3t+2)=0\2t[t(t^2+2t+1)+2(t+1)]=0\ 2t[(t(t+1)^2+2(t+1)]=0\ 2t[(t+1)(t^2+t)+2(t+1)]=0\ 2t(t+1)(t^2+t+2)=0\ t_1=0, t_2=-1[/latex]x-3 = 0 или х - 3 = -1х = 3    или   х = 2Ответ: 2; 3.