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Answer 1

Ответ:

Объяснение:

1) $\frac{5}{a-b}$ ·$\frac{a}{a}$ = $\frac{5*a}{a(a-b)}$= $\frac{5a}{a^{2}-ab }$

2) $\frac{4}{m+n}$ * $\frac{m+n}{m+n}$ = $\frac{4*(m+n)}{(m+n)(m+n)}$ = $\frac{4m+4n}{(m+n)^{2} }$ = $\frac{4m+4n}{m^{2} +2mn+n^{2} }$

3) $\frac{4}{k-1}$ * $\frac{k^{2} +k+1}{k^{2}+k+1 }$ = $\frac{4*(k^{2}+k+1) }{(k-1)(k^{2}+k+1) }$ =$\frac{4k^{2} +4k+4}{k^{3}-1 }$

Answer 2

$\displaystyle\bf\\1)\\\\\frac{5}{a-b} =\frac{5\cdot a}{(a-b)\cdot a} =\frac{5a}{a^{2} -ab} \\\\\\2)\\\\\frac{4}{m+n} =\frac{4\cdot(m+n)}{(m+n)\cxdot(m+n)}=\frac{4m+4n}{m^{2} +2mn+n^{2} } \\\\\\3)\\\\\frac{4}{k-1} =\frac{4\cdot(k^{2} +k+1)}{(k-1)\cdot(k^{2}+k+1) } =\frac{4k^{2} +4k+4}{k^{3}-1 }$