$(x-1{)}^4$ + $(x-3{)}^4$ =16 Tim x

2 bình luận về “$(x-1{)}^4$ + $(x-3{)}^4$ =16 Tim x”

1. (x-1)^4 + (x-3)^4  = 16 (1)

   Đặt x – 2= m

   (1) <=> ( m + 1)^4 + (m – 1)^4 = 16

   <=> m^4 + 4m^3 + 6m^2 + 4m + 1 + m^4 – 4m^3 + 6m^2 – 4m + 1 = 16

   <=> 2m^4 + 12m^2 + 2 = 16

   <=> m^4 + 6m^2 + 1 = 8

   <=> m^4 + 6m^2 – 7 = 0

   <=> m^4 – m^2 + 7m^2 – 7  = 0

   <=> m^2(m^2 – 1) + 7(m^2 – 1) = 0

   <=> (m^2 + 7)(m^2 – 1 ) = 0

   <=> (m^2 + 7)(m-1)(m+1) = 0

   <=> (m-1)(m+1) = 0 ( do m^2 + 7 ≥ 7 > 0 )

   <=> $[\begin{array}{l}m-1=0\\ m+1=0\end{array}$ 

   <=> $[\begin{array}{l}m=1\\ m=-1\end{array}$ 

   <=> $[\begin{array}{l}x=3\\ x=1\end{array}$ 

   Vậy x \in \{3;1\}

   Trả lời
2. Giải đáp:

   (x-1)^4 + (x-3)^4 = 16\ (*)

   Đặt x-2 = t ta có:

   (*) <=> (t+1)^4 + (t-1)^4 = 16

   <=> (t^4 + 4t^3 + 6t^2 + 4t + 1) + (t^4 – 4t^3 + 6t^2 – 4t + 1) = 16

   <=> t^4 + 4t^3 + 6t^2 + 4t + 1 + t^4 – 4t^3 + 6t^2 – 4t + 1 = 16

   <=> 2t^4 + 12t^2 + 2 = 16

   <=> 2(t^4 + 6t^2 + 1) = 16

   <=> t^4 + 6t^2 + 1 = 8

   <=> t^4 + 6t^2 – 7 = 0

   <=> (t^4 – t^2) + (7t^2 – 7) = 0

   <=> t^2 (t^2 – 1) + 7(t^2 -1)=0

   <=> (t^2 – 1)(t^2+7) = 0

   <=> $[\begin{array}{l}t=1\\ t=-1\end{array}$ 

   <=> $[\begin{array}{l}x=3\\ x=1\end{array}$ 

   Vậy x in {3;1}

   Trả lời