x2 + 4x + 3 = 0
\(\Leftrightarrow\)x2 + x + 3x + 3 = 0
\(\Leftrightarrow\)x(x + 1) + 3(x + 1) = 0
\(\Leftrightarrow\)(x + 1)(x + 3) = 0
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\)
Vậy....
\(x^2+4x+3=0\)
\(x^2+x+3x+3=0\)
\(\left(x^2+x\right)+\left(3x+3\right)=0\)
\(x\left(x+1\right)+3\left(x+1\right)=0\)
\(\left(x+1\right).\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}}\)
Vậy x=-1 hoặc x=-3
chọn mk nha
Ta có:
x2 +4x+3=0
x2+x+3x+3=0
(x2+x)+(3x+3)=0
x(x+1)+3(x+1)=0
(x+1)(x+3)=0
Suy ra: x+1=0 hoặc x+3=0
TH1:x+1=0 TH2: x+3=0
x=0-1 x=0-3
x=-1 x=-3
Vậy x=-1 hoặcx=-3
