Question

VA ROG !!!!!!!!! Ajutati dau coroana .Exercitiul 7b si 9a .Va rog imi trebuie urgent .Intrebare de viata si moarte .

Answer 1

7b)

$\it \dfrac{a}{-1+a^2}+\dfrac{1}{a^2-1}=\dfrac{a}{a^2-1}+\dfrac{1}{a^2-1}=\dfrac{a+1}{a^2-1}=\dfrac{a+1}{(a-1)(a+1)}=\dfrac{1}{a-1}$

9a)

$\it \dfrac{a^2-10a+25}{(a-4)^2-1}=\dfrac{(a-5)^2}{(a-4-1)(a-4+1)}=\dfrac{(a-5)(a-5)}{(a-5)(a-3)}=\dfrac{a-5}{a-3}$

Answer 2

$7) \\ \\ b) \frac{a}{ - 1 + {a}^{2} } + \frac{1}{ {a}^{2} - 1} = \frac{a}{ {a}^{2} - {1}^{2} } + \frac{1}{ {a}^{2} - 1 } = \frac{a}{(a + 1)(a - 1)} + \frac{1}{ {a}^{2} - {1}^{2} } = \frac{a}{(a + 1)(a - 1)} + \frac{1}{(a + 1)(a - 1)} = \frac{a + 1}{(a + 1)(a - 1) } = \frac{1}{a - 1}$

$9) \\ \\ a) \frac{ {a}^{2} - 10a + 25 }{ {(a - 4)}^{2} - 1 } = \frac{ {a}^{2} + 2a( - 5) + {( - 5)}^{2} }{ {(a - 4)}^{2} - {1}^{2} } = \frac{ {(a - 5)}^{2} }{((a - 4) - 1)((a - 4) + 1)} = \frac{ {(a - 5)}^{2} }{(a - 4 - 1)(a - 4 + 1)} = \frac{ {(a - 5)}^{2} }{(a - 5)(a - 3)} = \frac{ {(a - 5)}^{2 - 1} }{a - 3} = \frac{a - 5}{a - 3}$