Упр.39.9 ГДЗ Мерзляк Поляков 7 класс (Алгебра)
- Упростите выражение:
1) $$\frac{a-\frac{a^2}{a+1}}{a-\frac{a}{a+1}}$$;
2) $$\frac{a-\frac{6a-9}{a}}{1-\frac{3}{a}}$$;
3) $$1-\frac{1}{1-\frac{a}{1-\frac{1}{a+1}}}$$;
4) $$\frac{\frac{2a-b}{b}+1}{\frac{2a+b}{b}-1}+\frac{3-\frac{b}{a}}{\frac{3a}{b}-1}$$.
$$\frac{a-\frac{a^2}{a+1}}{a-\frac{a}{a+1}}= \frac{\frac{a(a+1)-a^2}{a+1}}{\frac{a(a+1)-a}{a+1}}= \frac{\frac{a}{a+1}}{\frac{a^2}{a+1}}= \frac{a}{a+1}\cdot\frac{a+1}{a^2}=\frac{1}{a}.$$
$$\frac{a-\frac{6a-9}{a}}{1-\frac{3}{a}}= \frac{\frac{a^2-(6a-9)}{a}}{\frac{a-3}{a}}= \frac{\frac{a^2-6a+9}{a}}{\frac{a-3}{a}}= \frac{\frac{(a-3)^2}{a}}{\frac{a-3}{a}}=a-3.$$
$$1-\frac{1}{1-\frac{1}{1-\frac{a}{a+1}}}= 1-\frac{1}{1-\frac{1}{\frac{a+1-a}{a+1}}}= 1-\frac{1}{1-\frac{1}{\frac{1}{a+1}}}= 1-\frac{1}{1-(a+1)}= 1-\frac{1}{-a}=1+\frac{1}{a}=\frac{a+1}{a}.$$
$$\frac{\frac{2a-b}{b}+1}{\frac{2a+b}{b}-1}+\frac{3-\frac{b}{a}}{\frac{3a}{b}-1} =\frac{\frac{2a-b+b}{b}}{\frac{2a+b-b}{b}}+\frac{\frac{3a-b}{a}}{\frac{3a-b}{b}}$$
$$=\frac{\frac{2a}{b}}{\frac{2a}{b}}+\frac{\frac{3a-b}{a}}{\frac{3a-b}{b}} =1+\frac{3a-b}{a}\cdot\frac{b}{3a-b} =1+\frac{b}{a} =\frac{a+b}{a}.$$
Ответ
1) $$\frac{1}{a}$$; 2) $$a-3$$; 3) $$\frac{a+1}{a}$$; 4) $$\frac{a+b}{a}$$.
