Question

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Answer 1

$\frac{1}{v + d} + \frac{3vd}{ {v}^{3} + {d}^{3} }$

Распишем v³+d³ по формуле a³ + b³ = (a + b)·(a² - ab + b²)

$\frac{1}{v + d} + \frac{3vd}{(v + d)( {v}^{2} - vd + {d}^{2}) }$

Возведём к общему знаменателю:

$\frac{1 \times ( {v}^{2} - vd + {d}^{2})}{(v + d)( {v}^{2} - vd + {d}^{2})} + \frac{3vd}{(v + d)( {v}^{2} - vd + {d}^{2}) } \\ \frac{{v}^{2} - vd + {d}^{2} + 3vd}{(v + d)( {v}^{2} - vd + {d}^{2})} \\ \frac{{v}^{2} + 2vd + {d}^{2} }{(v + d)( {v}^{2} - vd + {d}^{2})}$

По формуле v²+2vd+d² = (v+d)²

$\frac{ {(v + d)}^{2} }{(v + d)( {v}^{2} - vd + {d}^{2})}$

Сократим:

$\frac{v + d}{ {v}^{2} - vd + {d}^{2} }$

Ответ: (v+d)/(v²-vd+d²).