Question

РЕШИТЕ ПРОШУ ПЖ ДАМ МНОГО БАЛЛОВ!!!​

Answer 1

(p² - 2p)/(p² - 4p + 4) = p(p-2)/(p-2)² = p/(p-2)

(a²-4)/a * 1/(a-2) - (a+2)/a = (a-2)(a+2)/a * 1/(a-2) - (a+2)/a = (a+2)/a -(a+2)/a = 0

3. (3n/(n-4) - 6n/(n²-8n+16)) : (n-6)/(16-n²) + 24n/(n-4) = -3n

3n/(n-4) - 6n/(n² - 8n + 16) = (3n(n-4) - 6n)/(n-4)² = (3n² - 18n)/(n-4)² = 3n(n - 6)/(n-4)²

3n(n - 6)/(n-4)² : (n-6)/(16-n²) = 3n(n - 6)/(n-4)² * (4-n)(4+n)/(n-6) = -3n(n+4)/(n-4)

-3n(n+4)/(n-4)  + 24n/(n-4) = (24n - 3n² - 12n)/(n-4) = (12n - 3n²)/(n - 4) = -3n(n - 4)/(n-4) = -3n

Answer 2

$\displaystyle \tt 1). \ \ \frac{p^{2}-2p}{p^{2}-4p+4}=\frac{p(p-2)}{(p-2)^{2}}=\frac{p}{p-2};\\\\\\2). \ \ \frac{a^{2}-4}{a}\cdot\frac{1}{a-2}-\frac{a+2}{a}=\frac{(a-2)(a+2)}{a(a-2)}-\frac{a+2}{a}=\frac{a+2}{a}-\frac{a+2}{a}=0;$

$\displaystyle \tt 3). \ \ \bigg(\frac{3n}{n-4}-\frac{6n}{n^{2}-8n+16}\bigg):\frac{n-6}{16-n^{2}}+\frac{24n}{n-4}=\\\\{} \ \ =\bigg(\frac{3n}{n-4}-\frac{6n}{(n-4)^{2}}\bigg)\cdot\frac{(4-n)(4+n)}{n-6}+\frac{24n}{n-4}=\\\\{} \ \ =\frac{3n(n-4)-6n}{(n-4)^{2}}\cdot\frac{(4-n)(4+n)}{n-6}+\frac{24n}{n-4}=\\\\{} \ \ =\frac{3n(n-6)}{(n-4)^{2}}\cdot\frac{(4-n)(4+n)}{n-6}+\frac{24n}{n-4}=\\\\{} \ \ =-\frac{3n(4+n)-24n}{n-4}=-\frac{3n^{2}-12n}{n-4}=-\frac{3n(n-4)}{n-4}=-3n;$