a) x2 - 2x + 4x - 8 = 0
=> x.(x - 2) + 4.(x - 2) = 0
=> (x - 2).(x + 4) = 0
=> \(\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}}\)=> \(\orbr{\begin{cases}x=2\\x=-4\end{cases}}\)
b) x(x + 3) - 3x - 9 = 0
=> x.(x + 3) - 3.(x + 3) = 0
=> (x + 3).(x - 3) = 0
=> \(\orbr{\begin{cases}x+3=0\\x-3=0\end{cases}}\)=> \(\orbr{\begin{cases}x=-3\\x=3\end{cases}}\)
c) x2 - 6x + 5 = 0
=> x2 - x - 5x + 5 = 0
=> x.(x - 1) - 5.(x - 1) = 0
=> (x - 1).(x - 5) = 0
=> \(\orbr{\begin{cases}x-1=0\\x-5=0\end{cases}}\)=> \(\orbr{\begin{cases}x=1\\x=5\end{cases}}\)
1/\(x^2-2x+4x-8=0\)
=>\(x\left(x-2\right)+4\left(x-2\right)=0\)
=>\(\left(x-4\right)\left(x-2\right)=0\)
=>\(\orbr{\begin{cases}x-4=0\\x-2=0\end{cases}}\)=>\(\orbr{\begin{cases}x=4\\x=2\end{cases}}\)
2/\(x\left(x+3\right)-3x-9=0\)
=>\(x\left(x+3\right)-3\left(x+3\right)=0\)
=>\(\left(x-3\right)\left(x+3\right)=0\)
=>\(\orbr{\begin{cases}x-3=0\\x+3=0\end{cases}}\)=>\(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
3/\(x^2-6x+5=0\)
=>\(x^2-x-5x+5=0\)
=>\(x\left(x-1\right)-5\left(x-1\right)=0\)
=>\(\left(x-5\right)\left(x-1\right)=0\)
=>\(\orbr{\begin{cases}x-5=0\\x-1=0\end{cases}}\)=>\(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
mình nhầm câu a, chỗ x-4 sửa thành x+4 và => x=-4 chứ ko phải là 4
làm lại nè:
a) x2 - 2x + 4x - 8 = 0
=> x(x - 2) + 4(x - 2) = 0
=> (x - 2)(x + 4) = 0
=> \(\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}}\)=>\(\orbr{\begin{cases}x=2\\x=-4\end{cases}}\)
b) x(x + 3) - 3x - 9 = 0
=> x(x + 3) - 3(x + 3) = 0
=> (x - 3)(x +3) = 0
=> \(\orbr{\begin{cases}x-3=0\\x+3=0\end{cases}}\)=>\(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
c) x2 - 6x + 5 = 0
=> x2 - x - 5x + 5 = 0
=> x(x - 1) - 5(x - 1) = 0
=> (x - 5)(x - 1) = 0
=> \(\orbr{\begin{cases}x-5=0\\x-1=0\end{cases}}\)=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
1) Ta có: \(x^2-2x+4x-8=0\)
\(\Rightarrow x\left(x-2\right)+4\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}\)
Vậy x=2; x=-4
2) Ta có: x(x+3)-3x-9=0
=> x(x+3)-3(x+3)=0
=> (x+3)(x-3)=0
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=3\end{cases}}\)
Vậy x=-3; x=3
3) Ta có: \(x^2-6x+5=0\)
\(\Rightarrow x^2-x-5x+5=0\)
\(\Rightarrow x\left(x-1\right)-5\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=5\end{cases}}\)
Vậy x=1; x=5
a) x2 - 2x + 4x - 8 = 0
=> x(x - 2) + 4(x - 2) = 0
=> (x - 2)(x + 4) = 0
=> $$=>$$
b) x(x + 3) - 3x - 9 = 0
=> x(x + 3) - 3(x + 3) = 0
=> (x - 3)(x +3) = 0
=> $$=>$$
c) x2 - 6x + 5 = 0
=> x2 - x - 5x + 5 = 0
=> x(x - 1) - 5(x - 1) = 0
=> (x - 5)(x - 1) = 0
=> $$=> $$
