a, \(2\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy x = 2 hoặc x = -3
b, \(8x^3-50x=0\)
\(\Leftrightarrow2x\left(4x^2-25\right)=0\)
\(\Leftrightarrow2x\left(2x-5\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\2x-5=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\)
Vậy...
1.
a) \(2\left(x+3\right)-x^2-3x=0\)
\(2\left(x+3\right)-\left(x^2+3x\right)=0\)
\(2\left(x+3\right)-x\left(x+3\right)=0\)
\(\left(2-x\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2-x=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy x=2 hoặc x=-3
b) \(8x^3-50x=0\)
\(x\left(8x^2-50\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\8x^2-50=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=\dfrac{25}{4}=\left(\pm\dfrac{5}{2}\right)^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=\dfrac{-5}{2}\) hoặc \(x=\dfrac{5}{2}\)
tik mik nhé !!!