Question

Tìm x,biết a)x^2-4x-5=0

Answer 1

\(a,x^2-4x-5=0\)

\(\Rightarrow x^2-5x+x-5=0\)

\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x+1=0\end{matrix}\right.\)              \(\Rightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

Vậy \(x\in\left\{5;-1\right\}\)

Answer 2

$x^2 - 4x - 5 = x^2 + x - 5x - 5 = x(x+1) - 5(x+1) = (x-5)(x+1)$

Suy ra $x^2 - 4x - 5 = 0 \Leftrightarrow (x-5)(x+1) = 0$

$\Leftrightarrow x = -1$ hoặc $x = 5$.