Question

giải phương trình \(x^2+2x-3=0\)

Answer 1

Ta có : x² + 2x - 3 = 0 

<=> x² - x + 3x - 3 = 0

<=> x(x - 1) + 3(x - 1) = 0

<=> (x + 3)(x - 1) = 0

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=1\end{cases}}\)

Answer 2

\(x^2+2x-3=0\)

\(x^2-x+3x-3=0\)

\(x\left(x-1\right)+3\left(x-1\right)=0\)

\(\left(x+3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=1\end{cases}}\)