Question

Помогите пожалуйста с алгеброй(

Answer 1

1.

а)

$\frac{3a+2b}{5a-3b} =\frac{3*21+2*28}{5*21-3*28} =\frac{63+56}{105-84} =\frac{119}{21} =5\frac{2}{3}$

б)

$\frac{25a^{2}-4b^{2}}{5a-2b} =\frac{(5a-2b)(5a+2b)}{5a-2b} =5a+2b=5*2.4+2*2.8=12+5.6=17.6$

2.

а)

$\frac{7m+14n}{21} =\frac{7(m+2n)}{21} =\frac{m+2n}{3}$

б)

$\frac{3a-a^{2}+5a^{3}}{3-a+5a^{2}} =\frac{a(3+a+5a^{2})}{3-a+5a^{2}} =a$

в)

$\frac{5a+3u}{25a^{2}+30au+9u^{2}} =\frac{5a+3u}{(5a+3u)^{2}} =\frac{1}{5a+3u}$

г)

$\frac{p^{2}+9q^{2}+6pq}{2p+6q} =\frac{(p+3q)^{2}}{2(p+3q)} =\frac{p+3q}{2}$

3.

а)

$\frac{3a+7}{a-1} +\frac{7a-5}{a-1} =\frac{3a+7+7a-5}{a-1} =\frac{10a-2}{a-1}$

б)

$\frac{6a+7}{a-1} -\frac{4a+8}{a-1} -\frac{2a-1}{a-1} =\frac{6a+7-4a-8-2a+1}{a-1} =\frac{0}{a-1} =0$

в)

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

г)

$\frac{2b-3c}{b-5c} +\frac{c-3b}{5c-b} =\frac{2b-3c}{b-5c} -\frac{c-3b}{b-5c} =\frac{2b-3c-c+3b}{b-5c} =\frac{5b-4c}{b-5c}$