\(\left(x-2\right)^3-\left(x+5\right)\left(x^2-5x+25\right)=0\)
\(\Rightarrow x^3-6x^2+12x-8-x^3-125=0\)
\(\Rightarrow-6x^2+12x-133=0\)
\(\Rightarrow6x\left(2-x\right)=133\)
\(\Rightarrow\left[{}\begin{matrix}6x=133\\2-x=133\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{113}{6}\\x=-131\end{matrix}\right.\)
Vậy ...........................
⇔ (x-2)3- (x+5)3=0
=> (x-2-x-5) [(x-2)2+(x-2)(x-5)+ (x+5)2] =0
=> -7 . [(x-2)2+(x-2)(x-5)+ (x+5)2] =0
=> [(x-2)2+(x-2)(x-5)+ (x+5)2] =0
=> \(\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x-2\right)\left(x-5\right)=0\\\left(x+5\right)^2=0\end{matrix}\right.\)\(\left\{\left[{}\begin{matrix}x=2\\x=2\\x=-5\end{matrix}\right.hoặc}x=5\right\}\)
