Упр.23.26 ГДЗ Мерзляк Поляков 9 класс (Алгебра)

1) (m-n)/(m^(1/3)-n^(1/3))-(m+n)/(m^(1/3)+n^(1/3));
2) (1-a^(1/36))(1+a^(1/36)+a^(1/18))+(4-a^(1/6))/(2-a^(1/12));
3) (a^(1/4)-b^(1/4)):(b^(5/4)/a-b/a^(3/4));
4) (m^(5/2)-m^(3/2))/(m^3-m^(5/2))-(m^(1/2)+m)/(m^2+m^(3/2)).

1. $$\frac{m-n}{m^{1/3}-n^{1/3}}-\frac{m+n}{m^{1/3}+n^{1/3}}$$
   Представим разность и сумму кубов:
   $$m-n=(m^{1/3}-n^{1/3})(m^{2/3}+m^{1/3}n^{1/3}+n^{2/3}),$$
   $$m+n=(m^{1/3}+n^{1/3})(m^{2/3}-m^{1/3}n^{1/3}+n^{2/3}).$$
   Тогда
   $$\frac{m-n}{m^{1/3}-n^{1/3}}-\frac{m+n}{m^{1/3}+n^{1/3}}$$
   $$=m^{2/3}+m^{1/3}n^{1/3}+n^{2/3}-\left(m^{2/3}-m^{1/3}n^{1/3}+n^{2/3}\right)$$
   $$=2m^{1/3}n^{1/3}.$$

2. $$\left(1-a^{1/36}\right)\left(1+a^{1/36}+a^{1/18}\right)+\frac{4-a^{1/6}}{2-a^{1/12}}$$
   Так как
   $$\left(1-a^{1/36}\right)\left(1+a^{1/36}+a^{1/18}\right)=1-a^{1/12},$$
   а
   $$4-a^{1/6}=\left(2-a^{1/12}\right)\left(2+a^{1/12}\right),$$
   то
   $$\left(1-a^{1/36}\right)\left(1+a^{1/36}+a^{1/18}\right)+\frac{4-a^{1/6}}{2-a^{1/12}}$$
   $$=1-a^{1/12}+2+a^{1/12}=3.$$

3. $$\left(a^{1/4}-b^{1/4}\right):\left(\frac{b^{5/4}}{a}-\frac{b}{a^{3/4}}\right)$$
   Приведём знаменатель к общему знаменателю:
   $$\frac{b^{5/4}}{a}-\frac{b}{a^{3/4}}=\frac{b^{5/4}a^{3/4}-ab}{a\cdot a^{3/4}}=\frac{ab\left(b^{1/4}-a^{1/4}\right)}{a\cdot a^{3/4}}.$$
   Тогда
   $$\left(a^{1/4}-b^{1/4}\right):\left(\frac{b^{5/4}}{a}-\frac{b}{a^{3/4}}\right)$$
   $$=\left(a^{1/4}-b^{1/4}\right)\cdot \frac{a\cdot a^{3/4}}{ab\left(b^{1/4}-a^{1/4}\right)}$$
   $$=-\frac{a}{b}.$$

4. $$\frac{m^{5/2}-m^{3/2}}{m^3-m^{5/2}}-\frac{m^{1/2}+m}{m^2+m^{3/2}}$$
   Вынесем общий множитель:
   $$\frac{m^{3/2}(m-1)}{m^{5/2}(m^{1/2}-1)}-\frac{m^{1/2}(1+m^{1/2})}{m^{3/2}(m^{1/2}+1)}.$$
   Сократим:
   $$\frac{m-1}{m(m^{1/2}-1)}-\frac{1}{m}.$$
   Так как
   $$m-1=(m^{1/2}-1)(m^{1/2}+1),$$
   то
   $$\frac{m^{1/2}+1}{m}-\frac{1}{m}=\frac{m^{1/2}}{m}=m^{-1/2}.$$

Ответ

1) $$2m^{1/3}n^{1/3}$$
2) $$3$$
3) $$-\frac{a}{b}$$
4) $$m^{-1/2}$$