\(\left(x-4\right)^2=\left(x-4\right)^4\)
\(\Rightarrow\left(x-4\right)^4-\left(x-4\right)^2=0\)
\(\Rightarrow\left(x-4\right)^2.\left[\left(x-4\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-4\right)^2=0\\\left(x-4\right)^2-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\\left(x-4\right)^2=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=4\\x-4\in\left\{1;-1\right\}\end{cases}}\)
\(\Rightarrow x\in\left\{4;5;-4\right\}\)
Đoạn cuối cho sửa thành x = 4 ; 5 hoặc 3 nhé
Ta có: (x - 4)2 = (x - 4)4
=> (x - 4)2 [1 - (x - 4)2 ] = 0
=> (x - 4)2 (1 - x2 + 8x - 16) = 0
=> (x - 4)(-x2 + 8x - 15) = 0
=> (x - 4)(x - 3)(x - 5) = 0
=> x = 4 hoặc x = 3 hoăc x = 5
Vậy x = 4 , x = 3, x = 5
\(\left(x-4\right)^2=\left(x-4\right)^4\)
\(\Rightarrow\left(x-4\right)^4-\left(x-4\right)^2=0\)
\(\Rightarrow\left(x-4\right)^2.\left[\left(x-4\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-4\right)^2=0\\\left(x-4\right)^2-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=4\\x-4\in\left\{1;-1\right\}\end{cases}}\)
\(\Rightarrow x\in\left\{3;4;5\right\}\)
\(\left(x-4\right)^2-\left(x-4\right)^4=0\)
\(\left(x-4\right)^2-\left[\left(x-4\right)^2\right]^2=0\)
\(\left[x-4+\left(x-4\right)^2\right]\left[x-4-\left(x-4\right)^2\right]=0\)
\(\left(x-4+x^2-8x+16\right)\left(x-4-x^2+8x-16\right)=0\)
\(\left(x^2-7x+12\right)\left(9x-x^2-20\right)=0\)
\(-\left(x^2-7x+12\right)\left(x^2-9x+20\right)=0\)
\(-\left(x^2-3x-4x+12\right)\left(x^2-4x-5x+20\right)=0\)
\(-\left[x\left(x-3\right)-4\left(x-3\right)\right]\left[x\left(x-4\right)-5\left(x+4\right)\right]=0\)
\(-\left(x-3\right)\left(x-4\right)^2\left(x-5\right)=0\)
=> x-3=0=>x=3
(x-4)2=0=>x-4=0 => x=4
x-5=0=>x=5
Chọn t i c k cho mình nha
