Question

Даю 100!!!!!!!!!!!!!!!!!!!!! срочно!!!!!!!!!!!

Answer 1

$1) \ \dfrac{z^{2}-4z+16 }{16z^{2} -1} \cdot\dfrac{4z^{2} +z}{z^{3} +64} -\dfrac{z+4}{4z^{2} -z} =\\\\\\=\dfrac{z^{2} -4z+16}{(4z-1)(4z+1)} \cdot\dfrac{z(4z+1)}{(z+4)(z^{2} -4z+16)} -\dfrac{z+4}{z(4z-1)} =\\\\\\=\dfrac{z}{(4z-1)(z+4)} -\dfrac{z+4}{z(4z-1)} =\dfrac{z\cdot z-(z+4)\cdot(z+4)}{z(4z-1)(z+4)} =\\\\\\=\dfrac{z^{2} -z^{2}-8z-16 }{z(4z-1)(z+4)} =-\dfrac{8(z+2)}{z(4z-1)(z+4)} \\\\\\\\-\dfrac{8(z+2)}{z(4z-1)(z+4)} :\dfrac{7}{z^{2} +4z} =-\dfrac{8(z+2)}{z(4z-1)(z+4)} \cdot\dfrac{z(z+4)}{7}$=

$=\dfrac{8(z+2)}{7(1-4z)} \\\\\\\dfrac{8(z+2)}{7(1-4z)}-\dfrac{20z+13}{7-28z}= \dfrac{8(z+2)}{7-28z}-\dfrac{20z+13}{7-28z}= \dfrac{8z+16-20z-13}{7-28z} =\\\\\\=\dfrac{3-12z}{7-28z} =\dfrac{3(1-4z)}{7(1-4z)} =\boxed{\dfrac{3}{7}}$