Question

помоги пожалуйста срочно​

Answer 1

Объяснение:

1)

$1.\ \frac{1}{a-c} -\frac{3c^2}{a^3-c^3}-\frac{c}{a^2+ac+c^2}= \frac{1}{a-c} -\frac{3c^2}{(a-c)*(a^2+ac+c^2)}-\frac{c}{a^2+ac+c^2}= \\ =\frac{a^2+ac+c^2-3c^2*1-c*(a-c)}{(a-c)*(a^2+ac+c^2)} =\frac{a^2+ac+c^2-3c^2-ac+c^2}{(a-c)*(a^2+ac+c^2)}=\frac{a^2-c^2}{(a-c)*(a^2+ac+c^2)} =\\=\frac{(a-c)*(a+c)}{(a-c)*(a^2+ac+c^2)}=\frac{a+c}{a^2+ac+c^2}.$

$2.\ c+\frac{a^2}{a+c} =\frac{c*(a+c)+a^2}{a+c} =\frac{ca+c^2+a^2}{a+c}=\frac{a^2+ac+c^2}{a+c} .$

$3.\ \frac{a+c}{a^2+ac+c^2} *\frac{a^2+ac+c^2}{a+c} =1.$

Ответ: 1.

2)

$1.\ \frac{c}{a^3-c^3}*\frac{a^2+ac+c^2}{a+c} = \frac{c}{(a-c)*(a^2+ac+c^2)} *\frac{a^2+ac+c^2}{a+c}=\frac{c}{(a+c)*(a-c)}.\\ 2.\ \frac{1}{a-c}-\frac{c}{(a+c)*(a-c)}=\frac{a+c-c}{(a+c)*(a-c)} =\frac{a}{a^2-c^2} .\\ 3.\ 3a*\frac{a}{a^2-c^2}=\frac{3a^2}{a^2-c^2}.$

$4.\ \frac{3a^2}{a^2-c^2} -\frac{3c^2}{a^2-c^2} =\frac{3a^2-3c^2}{a^2-c^2}=\frac{3*(a^2-c^2)}{a^2-c^2}=3.$

Ответ: 3.