Ta có: \(x^3+12x^2+48x+64=8x^3-12x^2+6x-1\)
\(\Leftrightarrow\left(x+2\right)^3=\left(2x-1\right)^3\)
\(\Leftrightarrow\left(x+2\right)^3-\left(2x-1\right)^3=0\)
\(\Leftrightarrow\left[\left(x+2\right)-\left(2x-1\right)\right]\left[\left(x+2\right)^2+\left(x+2\right)\left(2x-1\right)+\left(2x-1\right)^2\right]=0\)
\(\Leftrightarrow\left(x+2-2x+1\right)\left(x^2+4x+4+2x^2+3x-2+4x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(3-x\right)\left(7x^2+3x-6\right)=0\)
\(\Leftrightarrow7\left(3-x\right)\cdot\left(x^2+\frac{3}{7}x-\frac{6}{7}\right)=0\)
mà 7>0
nên \(\left[{}\begin{matrix}3-x=0\\x^2+\frac{3}{7}x-\frac{6}{7}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x^2+2\cdot x\cdot\frac{3}{14}+\frac{9}{196}-\frac{177}{196}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\\left(x+\frac{3}{14}\right)^2=\frac{177}{196}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+\frac{3}{14}=\frac{\sqrt{177}}{14}\\x+\frac{3}{14}=-\frac{\sqrt{177}}{14}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-3+\sqrt{177}}{14}\\x=\frac{-3-\sqrt{177}}{14}\end{matrix}\right.\)
Vậy: \(S=\left\{3;\frac{-3+\sqrt{177}}{14};\frac{-3-\sqrt{177}}{14}\right\}\)
x3 + 12x2 + 48x + 64 = 8x3 - 12x2 + 6x - 1
\(\Leftrightarrow\) x3 + 12x2 + 48x + 64 - 8x3 + 12x2 - 6x + 1 = 0
\(\Leftrightarrow\) -7x3 + 24x2 + 42x + 65 = 0
Bn cho đề thế này ai mà giải được :vvv