Giải:
Ta có phương trình:
\(\dfrac{x^3+8}{2}=\left(\dfrac{x}{2}+1\right)^3\)
\(\Leftrightarrow\dfrac{x^3+8}{2}=\dfrac{x^3}{8}+\dfrac{3x^2}{4}+\dfrac{3x}{2}+1\)
\(\Leftrightarrow\dfrac{4\left(x^3+8\right)}{8}=\dfrac{x^3}{8}+\dfrac{2.3x^2}{8}+\dfrac{4.3x}{8}+\dfrac{8}{8}\)
\(\Leftrightarrow4x^3+32=x^3+6x^2+12x+8\)
\(\Leftrightarrow4x^3-x^3-6x^2-12x=8-32\)
\(\Leftrightarrow3x^3-6x^2-12x=-24\)
\(\Leftrightarrow3x^3-6x^2-12x+24=0\)
\(\Leftrightarrow3\left(x^3-2x^2-4x+8\right)=0\)
\(\Leftrightarrow x^3-2x^2-4x+8=0\)
\(\Leftrightarrow x^2\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)
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