Упр.23.18 ГДЗ Мерзляк Поляков 9 класс (Алгебра)

1) (a+2a^(1/3))/(a^(2/3)+2);
2) (m^(5/4) n^(1/4)-m^(1/4) n^(5/4))/(m^(5/4) n^(5/4));
3) (a-b^2)/(a-a^(1/2) b);
4) (a-b)/(a^(2/3)+a^(1/3) b^(1/3)+b^(2/3));
5) (a^0,5-b^0,5)/(a-b);
6) (x^3,5 y^2,5-x^2,5 y^3,5)/(x+2x^0,5 y^0,5+y);
7) (a-125)/(a^(2/3)-25);
8) (m^(7/6)-36m^(5/6))/(m^(1/2)-6m^(1/2));
9) (24^(1/4)-8^(1/4))/(6^(1/4)-2^(1/4)).

1. $$\frac{a+2a^{1/3}}{a^{2/3}+2}=\frac{a^{1/3}\left(a^{2/3}+2\right)}{a^{2/3}+2}=a^{1/3}.$$

2. $$\frac{m^{5/4}n^{1/4}-m^{1/4}n^{5/4}}{m^{5/4}n^{5/4}}= \frac{m^{1/4}n^{1/4}\left(m-n\right)}{m^{5/4}n^{5/4}}= \frac{m-n}{mn}=\frac1n-\frac1m.$$

3. $$\frac{a-b^2}{a-a^{1/2}b}= \frac{\left(a^{1/2}-b\right)\left(a^{1/2}+b\right)}{a^{1/2}\left(a^{1/2}-b\right)}= \frac{a^{1/2}+b}{a^{1/2}}=1+\frac{b}{a^{1/2}}.$$

4. $$\frac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}= \frac{\left(a^{1/3}-b^{1/3}\right)\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}= a^{1/3}-b^{1/3}.$$

5. $$\frac{a^{0,5}-b^{0,5}}{a-b}= \frac{\left(a^{1/2}-b^{1/2}\right)}{\left(a^{1/2}-b^{1/2}\right)\left(a^{1/2}+b^{1/2}\right)}= \frac1{a^{1/2}+b^{1/2}}.$$

6. $$\frac{x^{3,5}y^{2,5}-x^{2,5}y^{3,5}}{x+2x^{0,5}y^{0,5}+y}= \frac{x^{2,5}y^{2,5}(x-y)}{\left(x^{0,5}+y^{0,5}\right)^2}.$$

   Так как $$x-y=\left(x^{0,5}-y^{0,5}\right)\left(x^{0,5}+y^{0,5}\right),$$ то
   $$\frac{x^{3,5}y^{2,5}-x^{2,5}y^{3,5}}{x+2x^{0,5}y^{0,5}+y}= x^{2,5}y^{2,5}\cdot\frac{x^{0,5}-y^{0,5}}{x^{0,5}+y^{0,5}}.$$

7. $$\frac{a-125}{a^{2/3}-25}= \frac{a^{1/3}-5}{a^{2/3}-5^2}= \frac{\left(a^{1/3}-5\right)\left(a^{2/3}+5a^{1/3}+25\right)}{\left(a^{1/3}-5\right)\left(a^{1/3}+5\right)}= \frac{a^{2/3}+5a^{1/3}+25}{a^{1/3}+5}.$$

8. $$\frac{m^{7/6}-36m^{5/6}}{m^{1/2}-6m^{1/2}}= \frac{m^{5/6}\left(m^{1/3}-36\right)}{m^{1/2}\left(1-6\right)}.$$

   Упростим числитель:
   $$m^{1/3}-36=\left(m^{1/6}-6\right)\left(m^{1/3}+6m^{1/6}+36\right).$$
   Тогда
   $$\frac{m^{7/6}-36m^{5/6}}{m^{1/2}-6m^{1/2}}= m^{1/3}\left(m^{1/6}+6\right).$$

9. $$\frac{24^{1/4}-8^{1/4}}{6^{1/4}-2^{1/4}}= \frac{2\cdot 6^{1/4}-2\cdot 2^{1/4}}{6^{1/4}-2^{1/4}}= 2\cdot\frac{6^{1/4}-2^{1/4}}{6^{1/4}-2^{1/4}}=2.$$

Ответ

1. $$a^{1/3}$$
2. $$\frac1n-\frac1m$$
3. $$1+\frac{b}{a^{1/2}}$$
4. $$a^{1/3}-b^{1/3}$$
5. $$\frac1{a^{1/2}+b^{1/2}}$$
6. $$x^{2,5}y^{2,5}\cdot\frac{x^{0,5}-y^{0,5}}{x^{0,5}+y^{0,5}}$$
7. $$\frac{a^{2/3}+5a^{1/3}+25}{a^{1/3}+5}$$
8. $$m^{1/3}\left(m^{1/6}+6\right)$$
9. $$2$$