Упр.23.16 ГДЗ Мордкович Семенов 8 класс (Алгебра)
Упростите выражение:
- $$\frac{\sqrt{a}}{\sqrt{bc}}+\frac{\sqrt{b}}{\sqrt{ac}}$$;
- $$\frac{3}{\sqrt{c}-5}+\frac{2}{\sqrt{c}}$$;
- $$\frac{\sqrt{a}-1}{3\sqrt{a}-12}-\frac{\sqrt{a}-2}{2\sqrt{a}-8}$$;
- $$\frac{\sqrt{n}}{\sqrt{mk}}-\frac{\sqrt{m}}{\sqrt{nk}}$$;
- $$\frac{\sqrt{a}+2}{\sqrt{a}-1}-\frac{\sqrt{a}-3}{\sqrt{a}}$$;
- $$\frac{\sqrt{d}-2}{3\sqrt{d}+3}-\frac{3\sqrt{d}-4}{7\sqrt{d}+7}$$.
а)
$$\frac{\sqrt a}{\sqrt{bc}}+\frac{\sqrt b}{\sqrt{ac}}= \frac{\sqrt a\sqrt a+\sqrt b\sqrt b}{\sqrt{abc}}= \frac{a+b}{\sqrt{abc}}.$$
б)
$$\frac{3}{\sqrt c-5}+\frac{2}{\sqrt c}= \frac{3\sqrt c+2(\sqrt c-5)}{\sqrt c(\sqrt c-5)}= \frac{5\sqrt c-10}{\sqrt c(\sqrt c-5)}= \frac{5(\sqrt c-2)}{\sqrt c(\sqrt c-5)}.$$
в)
$$\frac{\sqrt a-1}{3\sqrt a-12}-\frac{\sqrt a-2}{2\sqrt a-8}= \frac{\sqrt a-1}{3(\sqrt a-4)}-\frac{\sqrt a-2}{2(\sqrt a-4)}$$
$$= \frac{2(\sqrt a-1)-3(\sqrt a-2)}{6(\sqrt a-4)}= \frac{2\sqrt a-2-3\sqrt a+6}{6(\sqrt a-4)}$$
$$= \frac{-\sqrt a+4}{6(\sqrt a-4)}= -\frac16.$$г)
$$\frac{\sqrt n}{\sqrt{mk}}-\frac{\sqrt m}{\sqrt{nk}}= \frac{\sqrt n\sqrt n-\sqrt m\sqrt m}{\sqrt{mnk}}= \frac{n-m}{\sqrt{mnk}}.$$
д)
$$\frac{\sqrt a+2}{\sqrt a-1}-\frac{\sqrt a-3}{\sqrt a}= \frac{\sqrt a(\sqrt a+2)-(\sqrt a-3)(\sqrt a-1)}{\sqrt a(\sqrt a-1)}$$
$$= \frac{a+2\sqrt a-(a-4\sqrt a+3)}{\sqrt a(\sqrt a-1)}= \frac{6\sqrt a-3}{\sqrt a(\sqrt a-1)}$$
$$= \frac{3(2\sqrt a-1)}{\sqrt a(\sqrt a-1)}.$$е)
$$\frac{\sqrt d-2}{3\sqrt d+3}-\frac{3\sqrt d-4}{7\sqrt d+7}= \frac{\sqrt d-2}{3(\sqrt d+1)}-\frac{3\sqrt d-4}{7(\sqrt d+1)}$$
$$= \frac{7(\sqrt d-2)-3(3\sqrt d-4)}{21(\sqrt d+1)}= \frac{7\sqrt d-14-9\sqrt d+12}{21(\sqrt d+1)}$$
$$= \frac{-2\sqrt d-2}{21(\sqrt d+1)}= \frac{-2(\sqrt d+1)}{21(\sqrt d+1)}= -\frac{2}{21}.$$
Ответ
а) $$\frac{a+b}{\sqrt{abc}}$$; б) $$\frac{5(\sqrt c-2)}{\sqrt c(\sqrt c-5)}$$; в) $$-\frac16$$; г) $$\frac{n-m}{\sqrt{mnk}}$$; д) $$\frac{3(2\sqrt a-1)}{\sqrt a(\sqrt a-1)}$$; е) $$-\frac{2}{21}$$.








