a) \(x^2+x-12=0\)
\(\left(a=1;b=1;c=-12\right)\)
\(\Delta=b^2-4ac=1-4.1.\left(-12\right)=49>0\)
* \(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-1+\sqrt{49}}{2}=3\)
* \(x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-1-\sqrt{49}}{2}=-4\)
b) \(20-3x^2-7x=0\)
(a = -3 ; b = -7; c = 20)
\(\Delta=b^2-4ac=\left(-7\right)^2-4.\left(-3\right).20=289>0\)
* \(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{7+\sqrt{289}}{-6}=-4\)
* \(x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{7-\sqrt{289}}{-6}=\dfrac{5}{3}\)
a) \(x^2+x-12=0\)
\(x^2-3x+4x-12=0\)
\(x\left(x-3\right)+4\left(x-3\right)=0\)
\(\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)
mk làm câu c nha .
c) ta có : \(x^3-19x-30=0\Leftrightarrow x^3+2x^2-2x^2-4x-15x-30=0\)
\(\Leftrightarrow x^2\left(x+2\right)-2x\left(x+2\right)-15\left(x+2\right)=0\Leftrightarrow\left(x^2-2x-15\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x^2+3x-5x-15\right)\left(x+2\right)=0\Leftrightarrow\left(x\left(x+3\right)-5\left(x+3\right)\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)\left(x+3\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+2=0\\x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\\x=-3\end{matrix}\right.\)
vậy \(x=5;x=-2;x=-3\)
