(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 = -1 х = 3    или   х = 2 Ответ: 2; 3.