Question

Упростите выражение
$\frac{9}{3y+6} + \frac{y^2-3y+2}{y^2-4}$

Answer 1

Объяснение:

$\frac{9}{3y + 6} + \frac{y{}^{2} - 3y + 2 }{y {}^{2} - 4} = \frac{9}{3(y + 2)} \times \frac{y {}^{2} - y - 2y + 2 }{(y - 2) \times (y + 2)} = \frac{3}{y + 2} \times \frac{y \times (y - 1) - 2(y - 1)}{(y - 2) \times (y + 2)} = \frac{3}{y + 2} + \frac{(y - 1) \times (y - 2)}{(y - 2) \times (y + 2)} = \frac{3}{y + 2} + \frac{y - 1}{y + 2} = \frac{3 + y - 1}{y + 2} = \frac{2 + y}{y + 2} = \frac{y + 2}{y + 2} = 1$

Answer 2

$\dfrac{9}{3y+6}+\dfrac{y^{2} -3y+2}{y^{2}-4 } =\dfrac{3}{y+2}+\dfrac{y^{2} -3y+2}{(y+2)(y-2) } =\dfrac{3\cdot(y-2)+y^{2}-3y+2 }{3(y-2)(y+2)}=\\\\\\=\dfrac{3y-6+y^{2}-3y+2 }{(y+2)(y-2)}=\dfrac{y^{2}-4 }{y^{2}-4 }=\boxed1$