\(\left(x+1\right)^3+\left(x-2\right)^3=\left(2x-1\right)^3\\ \Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8=8x^3-12x^2+6x-1\\ \Leftrightarrow2x^3-3x^2+15x-7-8x^3+12x^2-6x+1=0\)\(\Leftrightarrow-6x^3+9x^2+9x-6=0\\ \Leftrightarrow-3\left(2x^3-3x^2+3x-2\right)=0\\ \Leftrightarrow2x^3-3x^2+3x-2=0\\ \Leftrightarrow\left(2x^3-2\right)-\left(3x^2-3x\right)=0\\ \Leftrightarrow2\left(x^3-1\right)-3x\left(x-1\right)=0 \)
\(\Leftrightarrow2\left(x^2+x+1\right)\left(x-1\right)-3x\left(x-1\right)=0\\ \Leftrightarrow\left(2x^2+2x+2-3x\right)\left(x-1\right)=0\\ \Leftrightarrow\left(2x^2-x+2\right)\left(x-1\right)=0\\ \Leftrightarrow\left(2x^2-x+\dfrac{1}{8}+\dfrac{15}{8}\right)\left(x-1\right)=0\\ \Leftrightarrow\left[\left(2x^2-x+\dfrac{1}{8}\right)+\dfrac{15}{8}\right]\left(x-1\right)=0\\ \Leftrightarrow\left[2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)+\dfrac{15}{8}\right]\left(x-1\right)=0\\ \Leftrightarrow\left[2\left(x-\dfrac{1}{4}\right)^2+\dfrac{15}{8}\right]\left(x-1\right)=0\\ \Leftrightarrow x-1=0\left(\text{Vì }2\left(x-\dfrac{1}{4}\right)^2+\dfrac{15}{8}\ne0\right)\\ \)
\(\Leftrightarrow x=1\)
Vậy phương trình có nghiệm là \(x=1\)
\(\left(x+1\right)^3+\left(x-2\right)^3=\left(2x-1\right)^3\\ \Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8=8x^3-12x^2+6x-1\\ \Leftrightarrow2x^3-3x^2+15x-7-8x^3+12x^2-6x+1=0 \)\(\Leftrightarrow-6x^3+9x^2+9x-6=0\\ \Leftrightarrow-3\left(2x^3-3x^2-3x+2\right)=0\\ \Leftrightarrow\left(2x^3+2\right)-\left(3x^2+3x\right)=0\)
\(\Leftrightarrow2\left(x^3+1\right)-3x\left(x+1\right)=0\\ \Leftrightarrow2\left(x^2-x+1\right)\left(x+1\right)-3x\left(x+1\right)=0\\ \Leftrightarrow\left(2x^2-2x+2-3x\right)\left(x+1\right)=0\\ \Leftrightarrow\left(2x^2-5x+2\right)\left(x+1\right)=0\\ \Leftrightarrow\left(2x^2-4x-x+2\right)\left(x+1\right)=0\\ \Leftrightarrow\left[\left(2x^2-4x\right)-\left(x-2\right)\right]\left(x+1\right)=0\\ \Leftrightarrow\left[2x\left(x-2\right)-\left(x-2\right)\right]\left(x+1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\\x=-1\end{matrix}\right.\)
Vậy tập nghiệm phương trình là \(S=\dfrac{1}{2};2;-1\)
