\(x^2-5x+6=0\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ x^2-2x-3x+6=0\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ x\cdot\left(x-2\right)-3\cdot\left(x+2\right)=0\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \left(x-3\right)\cdot\left(x-2\right)=0\Rightarrow x\in\left(2,3\right)\)
\(x^2-7x+12=0\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ x^2-3x-4x+12=0\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ x\cdot\left(x-3\right)-4\left(x-3\right)=0\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \left(x-4\right)\cdot\left(x-3\right)=0\Rightarrow x\in\left(3,4\right)\)
\(x^2+x-20=0\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ x^2+5x-4x-20=0\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ x\cdot\left(x+5\right)-4\cdot\left(x+5\right)=0\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \left(x-4\right)\cdot\left(x+5\right)=0\Rightarrow x\in\left(4,-5\right)\)
câu 4 mk chịu
\(1.x^2-5x+6=0\\ x^2-2x-3x+6=0\\ \left(x^2-2x\right)+\left(-3x+6\right)=0\\ x\left(x-2\right)-3\left(x-2\right)=0\\ \left(x-2\right)\left(x-3\right)=0\\ \left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
\(2.x^2-7x+12=0\\ x^2-3x-4x+12=0\\ \left(x^2-4x\right)+\left(-3x+12\right)=0\\ x\left(x-4\right)-3\left(x-4\right)=0\\ \left(x-4\right)\left(x-3\right)=0\\ \left[{}\begin{matrix}x-4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)
\(3.x^2+x-20=0\\ x^2-4x+5x-20=0\\ \left(x^2-4x\right)+\left(5x-20\right)=0\\ x\left(x-4\right)+5\left(x-4\right)=0\\ \left(x-4\right)\left(x+5\right)=0\\ \left[{}\begin{matrix}x-4=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
câu 4 mik nghĩ là đề sai
sửa đề bài câu 4 ạ
4x\(^2\)+5x-6=0
