Question

Giải phương trình 
x² + 7x + 12 = 0 
giúp tuiiii

Answer 1

Ta có : x² + 7x + 12 = 0

=> x² + 4x + 3x + 12 = 0

=> x(x + 4) + 3(x + 4) = 0

=> (x + 3)(x + 4) = 0

=> \(\orbr{\begin{cases}x+3=0\\x+4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-3\\x=-4\end{cases}}\)

Vậy \(x\in\left\{-3;-4\right\}\)

Answer 2

x² + 7x + 12 = 0

<=> x² + 3x + 4x + 12 = 0

<=> x( x + 3 ) + 4( x + 3 ) = 0

<=> ( x + 3 )( x + 4 ) = 0

<=> \(\orbr{\begin{cases}x+3=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-4\end{cases}}\)

Vậy S = { -3 ; -4 }

Answer 3

\(x^2+7x+12=0\)

\(\Leftrightarrow x^2+4x+3x+12=0\)

\(\Leftrightarrow x\left(x+4\right)+3\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x+4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=-4\end{cases}}\)

Answer 4

-.-

Ta có: \(x^2+7x+12=0\)

\(\Leftrightarrow\left(x^2+7x+\frac{49}{4}\right)-\frac{1}{4}=0\)

\(\Leftrightarrow\left(x+\frac{7}{2}\right)^2-\left(\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow\left(x+\frac{7}{2}-\frac{1}{2}\right)\left(x+\frac{7}{2}+\frac{1}{2}\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=-4\end{cases}}\)

Answer 5

x ² +7x+12=0 

X²+7X=-12

\(\Rightarrow\)\(\orbr{\begin{cases}X^2=0\\7X=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}X=0\\X=0\end{cases}}\)

Answer 6

\(x^2+3x+4x+12=0\)    

\(x\left(x+3\right)+4\left(x+3\right)=0\)    

\(\left(x+3\right)\left(x+4\right)=0\)   

\(\orbr{\begin{cases}x+3=0\\x+4=0\end{cases}}\)             

\(\orbr{\begin{cases}x=-3\\x=-4\end{cases}}\)