Sửa đề:
\(x^2-x=12\\\Leftrightarrow x^2-x-12=0\\ \Leftrightarrow x^2+3x-4x-12=0\\\Leftrightarrow x\left(x+3\right)-4\left(x+3\right)=0\\\Leftrightarrow \left(x-4\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{4;-3\right\}\)
\(\Leftrightarrow\)-x . ( x + 1 ) = 12
\(\Leftrightarrow\frac{-1}{x}=\frac{12}{x+1}\)
\(\Leftrightarrow\frac{-x-1}{x.\left(x+1\right)}=\frac{12x}{x.\left(x+1\right)}\)
\(\Leftrightarrow-x-1=12x\)
\(\Leftrightarrow-x-1-12x=0\)
\(\Leftrightarrow-13x-1=0\)
\(\Leftrightarrow x=-\frac{1}{13}\)
Sửa đề: \(x^2-x=12\)
\(\Leftrightarrow x^2-x-12=0\)
\(\Leftrightarrow x^2-4x+3x-12=0\)
\(\Leftrightarrow x\left(x-4\right)+3\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;4\right\}\)
