Question

4. доведіть тотожність:​

Answer 1

Ответ:

a) $\frac{a+2}{a^2+3a}-\frac{2}{3a}=\frac{1}{3a+9}$

б) $\frac{a-2}{a+1}+\frac{a+1}{a-1}-\frac{1}{1+a}\neq \frac{2a^2+4}{a^2-1}$

Объяснение:

a)

$\frac{a+2}{a^2+3a}-\frac{2}{3a}=\frac{a+2}{a(a+3)}-\frac{2}{3a}=\frac{3(a+2)}{3a(a+3)}-\frac{2(a+3)}{3a(a+3)}=\\\\\frac{3a+6-2a-6}{3a(a+3)}=\frac{a}{3a(a+3)}=\frac{1}{3(a+3)}=\frac{1}{3a+9}$

б)

$\frac{a-2}{a+1}+\frac{a+1}{a-1}-\frac{1}{1+a}=\frac{(a-2)(a-1)}{(a+1)(a-1)}+\frac{(a+1)(a+1)}{(a+1)(a-1)}-\frac{a-1}{(a+1)(a-1)}=\\\\\frac{(a-2)(a-1)+(a+1)^2-(a-1)}{a^2-1}=\frac{a^2-a-2a+2+a^2+2a+1-a+1}{a^2-1}=\frac{2a^2-2a+4}{a^2-1}$