Question

(4a-1/a²+4a + 4a+1/a²-4a) × a²-16/a²+1 при а=5 в -1 степени.

Answer 1

$\Big(\dfrac{4a-1}{a^{2}+4a }+\dfrac{4a+1}{a^{2}-4a } \Big)\cdot\dfrac{a^{2}-16 }{a^{2}+1 }=\Big(\dfrac{4a-1}{a(a+4) }+\dfrac{4a+1}{a(a-4) } \Big)\cdot\dfrac{a^{2}-16 }{a^{2}+1 }=\\\\\\=\dfrac{(4a-1)\cdot(a-4)+(4a+1)\cdot(a+4)}{a(a+4)(a-4)}\cdot\dfrac{(a+4)(a-4)}{a^{2}+1 } =\\\\\\=\dfrac{4a^{2} -16a-a+4+4a^{2} +16a+a+4}{a^{2}+1 }=\dfrac{8a^{2}+8 }{a^{2}+1 }=\dfrac{8\cdot(a^{2}+1) }{a^{2}+1 }=\boxed8$