Помогите решить! Буду благодарен.См.скрин

$[tex] \frac{m-8}{5m} / \frac{m^{2}-64}{15m^{2}}= \frac{m-8}{5m} * \frac{15m^{2}}{(m-8)(m+8)} = \frac{3m}{m+8}; \\( \frac{a^{-3}b^{2}}{2c} ) ^{2}* \frac{2c^{3} a^{6}}{b^{3}}= \frac{b^{4}2c^{3}a^{6}}{a^{6}4c^{2}b^{3}}= \frac{bc}{2}$ [/tex]

$64x+x^{-1}=-16; 64x+ \frac{1}{x} +16=0; 64x^{2}+1+16x=0; (8x+1)^{2}=0; 8x+1=0; x=- \frac{1}{8}$

$\frac{ 2^{5} 0,5 ^{-6} }{16 ^{3} }= \frac{ 2^{5} 10 ^{6} }{ 5^{6} (2*8) ^{3} } = \frac{2^{5} (2*5) ^{6}}{5^{6} 2 ^{12} }= \frac{2^{6} 5 ^{6}}{5^{6} 2 ^{12} }= \frac{1}{ 2^{6}} = 2^{-6}$

$( \frac{c-2}{c+2}- \frac{c}{c-2} )* \frac{c+2}{2-3c}= \frac{ c^{2} -4c+4- c^{2}-2c }{(c-2)(c+2)} * \frac{c+2}{2-3c}= \frac{-6c+4}{(c-2)(2-3c)} = \frac{2}{c-2}$