Question

СРОЧНО ПОМОГИТЕ ДАЮ 40 БАЛЛОВ

Answer 1

$\displaystyle\bf\\1)\\\\\frac{6z+1}{z-3} +\frac{6z-1}{z+3} =\frac{(6z+1)\cdot(z+3)+(6z-1)\cdot(z-3)}{(z-3)\cdot(z+3)} =\\\\\\=\frac{6z^{2}+18z+z+3+6z^{2} -18z-z+3 }{(z-3)\cdot(z+3)} =\frac{12z^{2}+6 }{(z-3)(z+3)}=\\\\\\=\frac{6\cdot(2z^{2}+1) }{z^{2} -9} \\\\\\2)\\\\\frac{z^{2}-9 }{2z^{2} +1} \cdot\frac{6\cdot(2z^{2} +1)}{z^{2} -9} =6\\\\\\Otvet \ : \ 6$