\(3x^3-48x=8\)
\(3x\left(x^2-16\right)=0\)
\(3x\left(x-4\right)\left(x+4\right)=0\)
\(\left[\begin{array}{nghiempt}x=0\\x-4=0\\x+4=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=0\\x=4\\x=-4\end{array}\right.\)
\(x^2-2x=24\)
\(x^2-2x-24=0\)
\(x^2-6x+4x-24=0\)
\(x\left(x-6\right)+4\left(x-6\right)=0\)
\(\left(x+4\right)\left(x-6\right)=0\)
\(\left[\begin{array}{nghiempt}x+4=0\\x-6=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=-4\\x=6\end{array}\right.\)
a)\(3x^3-48x=0\\ \Rightarrow3x\left(x^2-16\right)=0\\ \Rightarrow3x\left(x-4\right)\left(x+4\right)=0\\ \Rightarrow\left\{{}\begin{matrix}3x=0\\x-4=0\\x+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
Vậy x\(\in\){0;\(\pm\)4}
b)\(x^2-2x=24\\ \Rightarrow x^2-2x-24=0\\ \Rightarrow x^2-6x+4x-24=0\\ \Rightarrow x\left(x-6\right)+4\left(x-6\right)=0\\ \Rightarrow\left(x-6\right)\left(x+4\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x-6=0\\x+4=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
Vậy x\(\in\){6;\(-\)4}
a) 3x3 - 48x = 0
⇔3x.(x2 - 16) = 0
⇔3x.(x2 - 42) = 0
⇔3x.(x-4).(x+4) = 0
⇔\(\left[{}\begin{matrix}3x=0\\x-4=0\\x+4=0\end{matrix}\right.\)\(\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
Còn câu b) bạn tự làm nha.
