Question

вычислите разность:
$\frac{9x { }^{2} + 16}{27x {}^{3} + 64 } - \frac{1}{3x + 4}$
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Answer 1

$= \frac{9x^{2}+16 }{(3x+4)(9x^{2}-12x+16) } - \frac{1}{3x+4} =$

$= \frac{9x^{2} +16-(9x^{2} -12x+16)}{(3x+4)*(9x^{2} -12x+16)} =$

$= \frac{9x^{2}+16-9x^{2} +12x-16 }{27x^{3}+64 } =$

$= \frac{12}{27x^{3} +64}$