Question

[Nguyễn Thị Như Quỳnh]

tìm x biet;

1) x² - 2x + 4x -8 = 0

2) x(x+3)-3x -9 =0

3) x² - 6x + 5 = 0

các ban giai giup minh nha

Answer 1

[soyeon_Tiểu bàng giải]

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

=> \(\orbr{\begin{cases}x-1=0\\x-5=0\end{cases}}\)=> \(\orbr{\begin{cases}x=1\\x=5\end{cases}}\)

Answer 2

[Trà My]

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/\(x^2-6x+5=0\)

=>\(x^2-x-5x+5=0\)

=>\(x\left(x-1\right)-5\left(x-1\right)=0\)

=>\(\left(x-5\right)\left(x-1\right)=0\)

=>\(\orbr{\begin{cases}x-5=0\\x-1=0\end{cases}}\)=>\(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)