Question

ПРОШУ ПОМОЩИ У ЛЮДЕЙ ПОНИМАЮЩИХ АЛГЕРБУ, ТЕ КТО НЕ В КУРСЕ, НЕ ЛЕЗЬТЕ!

Answer 1

Объяснение:

1) $\frac{a-1}{2a-24} + \frac{a}{a-12} = \frac{a-1}{2(a-12)} + \frac{a}{a-12} = \frac{a-1}{2(a-12)} + \frac{2a}{2(a-12)} = \frac{a-1+2a}{2(a-12)} = \frac{3a-1}{2(a-12)}$

2) $(\frac{a}{d} - \frac{d}{a} )\frac{6ad}{a+d} = (\frac{a*a}{a*d} - \frac{d*d}{a*d} )\frac{6ad}{a+d} = (\frac{a^2-d^2}{a*d} )\frac{6ad}{a+d} = \frac{(a^2-d^2)6ad}{(a+d)(a*d)} = \frac{(a-d)(a+d)6ad}{(a+d)(a*d)} = \frac{(a-d)6ad}{(a*d)} = (a-d)6$

Answer 2

Решение и ответ:

$\displaystyle \frac{{a-1}}{{2a-24}}+\frac{a}{{a-12}}=\frac{{a-1}}{{2(a-12)}}+\frac{a}{{a-12}}= \frac{{a-1}}{{2(a-12)}}+\frac{{2a}}{{2(a-12)}}=\frac{{a-1+2a}}{{2(a-12)}}= \boxed{\frac{{3a-1}}{{2(a-12)}}}$

$\displaystyle \left( {\frac{a}{d} - \frac{d}{a}} \right) \cdot \frac{{6ad}}{{a+d}}=\left( {\frac{{aa}}{{ad}} - \frac{{dd}}{{da}}} \right) \cdot \frac{{6ad}}{{a+d}}=\left( {\frac{{{a^2}-{d^2}}}{{ad}}} \right) \cdot \frac{{6ad}}{{a+d}}=\frac{{6(a-d)(a+d)}}{{a+d}}=\boxed{6(a-d)}$