a) ta có : \(\left(x-1\right)^3+3x\left(x-4\right)+1=0\)
\(\Leftrightarrow x^3-3x^2+3x-1+3x^2-12x+1=0\)
\(\Leftrightarrow x^3-9x=0\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\\x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\) vậy \(x=0;x=\pm3\)
b) \(x^2-25=6x-9\Leftrightarrow x^2-6x-16=0\)
\(\Leftrightarrow x^2+2x-8x-16=0\Leftrightarrow x\left(x+2\right)-8\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
vậy \(x=8;x=-2\)
a )
\(\left(x-1\right)^3+3x\left(x-4\right)+1=0\)
\(\Leftrightarrow x^3-3x^2+3x-1+3x^2-12x+1=0\)
\(\Leftrightarrow x^3-9x=0\)
\(\Leftrightarrow x\left(x^2-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-9=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm3\end{matrix}\right.\)
Vậy \(x\in\left\{0;\pm3\right\}\)
b )
\(x^2-25=6x-9\)
\(\Leftrightarrow x^2-6x=25-9\)
\(\Leftrightarrow x^2-6x=16\)
\(\Leftrightarrow x^2-6x-16=0\)
\(\Leftrightarrow x^2-2.x.3+9-25=0\)
\(\Leftrightarrow\left(x-3\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
Vậy ...
