\(2x^2-3x-5=0\\ 2x^2+2x-5x-5=0\\ 2x\left(x+1\right)-5\left(x+1\right)=0\\ \left(x+1\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{5}{2}\end{matrix}\right.\\ \)
vậy ...
2x^2 - 3x - 5 = 0
<=> x^2 - 3x/2 - 5/2 = 0
<=> x^2 + x - 5/2x - 5/2 = 0
<=> (x +1) (x - 5/2)= 0
<=> x=-1 hoặc x = 5/2
\(\Delta=\left(-3\right)^2-4.\left(-5\right).2=9+40=49>0\)
\(\Delta>0\) thì pt có 2 nghiệm phân biệt:
\(\left\{{}\begin{matrix}x_1=\dfrac{3+7}{4}=\dfrac{10}{4}=\dfrac{5}{2}\\x_2=\dfrac{3-7}{4}=-1\end{matrix}\right.\)
\(Ta\) \(có\) \(2x^2-3x-5=0\)
\(\Rightarrow\) \(2x^2+2x-5x-5=0\)
\(\Rightarrow\) \(2x.\left(x+1\right)-5.\left(x+1\right)=0\)
\(\Rightarrow\) \(\left(x+1\right).\left(2x-5\right)=0\)
\(\Rightarrow\) \(\left[{}\begin{matrix}x+1=0\\2x-5=0\end{matrix}\right.\) \(\Rightarrow\)\(\left[{}\begin{matrix}x=-1\\x=\dfrac{5}{2}\end{matrix}\right.\)
\(Vậy\) \(đa\) \(thức\) \(2x^2-3x-5\) \(có\) \(nghiệm\) \(x=-1\) \(hoặc\) \(x=\dfrac{5}{2}\)
