Question

15 Simplificati:
$a) \: \: \frac{x {}^{2} - 1 }{x { }^{2} + x - 2}$
$b) \: \: \: \frac{x {}^{2} - 4}{x {}^{2} - 3 x - 10 }$
$c) \: \: \: \frac{x {}^{2} - 7x + 12}{x {}^{2} - 9 }$
$d) \: \: \: \frac{x {}^{2} - 16}{x {}^{2} - 3x - 4}$
​

Answer 1

Răspuns:

$a)\:\frac{x^{2}-1 }{x^{2} +x-2} =\frac{(x-1)(x+1)}{x^{2} +2x-x-2} = \frac{(x-1)(x+1)}{x(x+2)-(x+2)}= \frac{(x-1)(x+1)}{(x+2)(x-1)}^{(x-1} =\frac{x-1}{x+2}$

$b) \: \frac{x^{2} -4}{x^{2} -3x-10} =\frac{(x-2)(x+2)}{x^{2} +2x-5x-10} = \frac{(x-2)(x+2)}{x(x+2)-5(x+2)}=\frac{(x-2)(x+2)}{(x+2)(x-5)}^{(x+2} =\frac{x-2}{x-5}$

$c) \: \frac{x^{2}-7x+12 }{x^{2}-9 } = \frac{x^{2}-3x-4x+12 }{(x-3)(x+3)} =\frac{x(x-3)-4(x-3)}{(x-3)(x+3)}=\frac{(x-3)(x-4)}{(x-3)(x+3))} ^{(x-3} =\frac{x-4}{x+3}$

$d) \: \frac{x^{2}-16 }{x^{2} -3x-4} =\frac{(x-4)(x+4)}{x^{2} +x-4x-4} = \frac{(x-4)(x+4)}{x(x+1)-4(x+1)} =\frac{(x-4)(x+4)}{(x+1)(x-4)}^{(x-4} =\frac{x+4}{x+1}$