a) x2 - 9 + (x + 3) = 0
=> (x - 3).(x + 3) + (x + 3) = 0
=> (x + 3).(x - 3 + 1) = 0
=> (x + 3).(x - 2) = 0
=> \(\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}}\)=> \(\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
b) x2 - 5x + 6 = 0
=> x2 - 2x - 3x + 6 = 0
=> x.(x - 2) - 3.(x - 2) = 0
=> (x - 2).(x - 3) = 0
=> \(\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\)=> \(\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
\(x^2-9+\left(x+3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x+3\right)+\left(x+3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}}\)
\(x^2-5x+6=0\)
\(\Rightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}}\)
a)x2-9+(x+3)=0
(x2-32)+(x+3)=(x-3)(x+3)+(x+3)=0
(x+3)(x-3+1)=0
(x+3)(x-2)=0
Do ddos x+3=0 hoặc x-2=0
=>x=-3 hoặc x=2
b)x2-5x+6=0
x(x-5)=0-6=-6
Ko có x thỏa mãn. Bạn xét x là ước của -6 thì x-5=(-6):x, sau đó cộng (-6):x với 5 thì x ko nhận giá trị ban đầu
a) \(x^2-9+\left(x+3\right)=0\)
\(\left(x^2-9\right)+\left(x+3\right)=0\)
\(\left(x+3\right)\left(x-3\right)+\left(x+3\right)=0\)
\(\left(x+3\right)\left(x-3+1\right)=0\)
\(\left(x+3\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
b) \(x^2-5x+6=0\)
\(x^2-2x-3x+6=0\)
\(\left(x^2-2x\right)-\left(3x-6\right)=0\)
\(x\left(x-2\right)-3\left(x-2\right)=0\)
\(\left(x-3\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)