phân tích tiếp:
x^2-6x+5x-30
= x(x-6)+5(x-6)
=(x-6)(x+5)
chúc bạn học tốt ^^
\(x^2-x-30=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{121}{4}=\left(x-\dfrac{1}{2}\right)^2-\dfrac{121}{4}=0\)Suy ra \(\left(x-\dfrac{1}{2}\right)^2=\dfrac{121}{4}\Rightarrow x-\dfrac{1}{2}\in\left\{\dfrac{11}{2};-\dfrac{11}{2}\right\}\Rightarrow x\in\left\{6;-5\right\}\)
\(x^2-x-30=0\\ \Leftrightarrow x^2-6x+5x-30=0\\ \Leftrightarrow\left(x^2-6x\right)+\left(5x-30\right)=0\\ \Leftrightarrow x\left(x-6\right)+5\left(x-6\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)
Vậy \(x=-5\) hoặc \(x=6\)
x2 - x - 30 =0
x2 - 52 - x - 5 = 0
( x - 5)( x + 5) -( x + 5) =0
( x + 5)(x - 6) =0
Vậy , x có hai ngiệm là { -5 ; 6}
