Question

rut gon phan thuc:

1 \(\dfrac{x^2-18x-19}{x^2-1}\)

2 \(\dfrac{x\left(4x^2-8x+4\right)}{2x^3-2x^2}\)

Answer 1

1) \(\dfrac{x^2-18x-19}{x^2-1}=\dfrac{x^2-19x+x-19}{\left(x-1\right)\left(x+1\right)}=\dfrac{x\left(x-19\right)+x-19}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x-19\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-19}{x-1}\)

2) \(\dfrac{x\left(4x^2-8x+4\right)}{2x^3-2x^2}=\dfrac{4x\left(x^2-2x+1\right)}{2x^2\left(x-1\right)}=\dfrac{4x\left(x-1\right)^2}{2x^2\left(x-1\right)}=\dfrac{2\left(x-1\right)}{x}\)

Answer 2

1.=\(\dfrac{(x^2+x)-(19x+19)}{(x+1)(x-1)}\)

=\(\dfrac{x(x+1)-19(x+1)}{(x+1)(x-1)}\)

=\(\dfrac{(x+1)(x-19)}{(x+1)(x-1)}\)

=\(\dfrac{x-19}{x-1}\)