Упр.14.12 ГДЗ Мордкович Семенов 8 класс (Алгебра)
а) 4m/(m + 2) — (4 + 8m + 3m^2)/(m^2 + 4m + 4);
б) (5 + 13p — 6p^2)/(9p^2 + 6p + 1) + 2p/(3p + 1);
в) (8y^2 — 9xy + x^2)/(x — y)^2 — 9y/(y — x);
г) (7n^2 + mn — 8m^2)/(m^2 — 2mn + n^2) — 8m/(n — m).
а) $$\frac{4m}{m+2}-\frac{4+8m+3m^2}{m^2+4m+4}$$
Так как $$m^2+4m+4=(m+2)^2,$$ получаем
$$ \frac{4m}{m+2}-\frac{4+8m+3m^2}{(m+2)^2} = \frac{4m(m+2)-(4+8m+3m^2)}{(m+2)^2} $$
$$ =\frac{4m^2+8m-4-8m-3m^2}{(m+2)^2} =\frac{m^2-4}{(m+2)^2} =\frac{(m-2)(m+2)}{(m+2)^2} =\frac{m-2}{m+2}. $$
б) $$\frac{5+13p-6p^2}{9p^2+6p+1}+\frac{2p}{3p+1}$$
Так как $$9p^2+6p+1=(3p+1)^2,$$ то
$$ \frac{5+13p-6p^2}{(3p+1)^2}+\frac{2p}{3p+1} = \frac{5+13p-6p^2+2p(3p+1)}{(3p+1)^2} $$
$$ =\frac{5+13p-6p^2+6p^2+2p}{(3p+1)^2} =\frac{15p+5}{(3p+1)^2} =\frac{5(3p+1)}{(3p+1)^2} =\frac{5}{3p+1}. $$
в) $$\frac{8y^2-9xy+x^2}{(x-y)^2}-\frac{9y}{y-x}$$
Поскольку $$y-x=-(x-y),$$ имеем
$$ \frac{8y^2-9xy+x^2}{(x-y)^2}-\frac{9y}{y-x} = \frac{8y^2-9xy+x^2-9y(y-x)}{(y-x)^2} $$
$$ =\frac{8y^2-9xy+x^2-9y^2+9xy}{(y-x)^2} =\frac{x^2-y^2}{(y-x)^2} =\frac{(x-y)(x+y)}{(x-y)^2} =\frac{x+y}{x-y}. $$
г) $$\frac{7n^2+mn-8m^2}{m^2-2mn+n^2}-\frac{8m}{n-m}$$
Так как $$m^2-2mn+n^2=(n-m)^2,$$ то
$$ \frac{7n^2+mn-8m^2}{(n-m)^2}-\frac{8m}{n-m} = \frac{7n^2+mn-8m^2-8m(n-m)}{(n-m)^2} $$
$$ =\frac{7n^2+mn-8m^2-8mn+8m^2}{(n-m)^2} =\frac{7n^2-7mn}{(n-m)^2} =\frac{7n(n-m)}{(n-m)^2} =\frac{7n}{n-m}. $$
Ответ
а) $$\frac{m-2}{m+2}$$; б) $$\frac{5}{3p+1}$$; в) $$\frac{x+y}{x-y}$$; г) $$\frac{7n}{n-m}$$.
