(x-1)(x+5) / (x-1)(2x+6) = 1
=> (x-1)(x+5) = (x-1)(2x+6)
=> \(x^2+5x-x-5=2x^2+6x-2x-6\)
=> \(x^2-2x^2+4x-4x-5+6=0\)
=> \(-x^2+1=0\)
=> \(-x^2=-1\)
=> \(x^2=1\)
=> x thuộc {-1; 1}
\(\frac{\left(x-1\right)\left(x+5\right)}{\left(x-1\right)\left(2x+6\right)}\)
đề vầy à
\(\frac{\left(x-1\right)\left(x+5\right)}{\left(x-1\right)\left(2x+6\right)}\) = 1 => (x-1)(x+5)=(x-1)(2x+6)
Mà x-1=x-1 => x+5=2x+6
=> 5-6= 2x-x
=> -1=x. Vậy x=-1