Question

Giải phương trình:

\(\left(3x+1\right)^3+\left(x-7\right)^3=8\left(2x-3\right)^{^{ }3}\)    

Answer 1

pt <=> \(\left(3x+1+x-7\right)\left(\left(3x+1\right)^2+\left(x-7\right)^2\right)=\left(4x-6\right)^3\)

\(\Leftrightarrow\left(4x-6\right)\left(9x^2+6x+1+x^2-14x+49-\left(4x-6\right)^2\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(10x^2-8x+50-16x^2+48x-36\right)=0\)

\(\orbr{\begin{cases}2x-3=0\\-6x^2+40x+14=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\-3x^2+20x+7=0\left(\cdot\right)\end{cases}}\)

pt(*) <=> (3x-1)(x+7)=0 <=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=-7\end{cases}}\)

Vậy x=...