Упр.10.24 ГДЗ Мерзляк 10 класс Базовый уровень (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^(1/3));
9) (24^(1/4)-8^(1/4))/(6^(1/4)-2^(1/4)).
$$\frac{a+2a^{1/3}}{a^{2/3}+2}=\frac{a^{1/3}\left(a^{2/3}+2\right)}{a^{2/3}+2}=a^{1/3}.$$
$$\frac{m^{5/4}n^{1/4}-m^{1/4}n^{5/4}}{m^{5/4}n^{5/4}}=\frac{m^{1/4}n^{1/4}\left(m-n\right)}{m^{5/4}n^{5/4}}=\frac{m-n}{mn}=\frac1n-\frac1m.$$
$$\frac{a-b^2}{a-a^{1/2}b}=\frac{\left(a^{1/2}-b\right)\left(a^{1/2}+b\right)}{a^{1/2}\left(a^{1/2}-b\right)}=\frac{a^{1/2}+b}{a^{1/2}}=1+\frac{b}{a^{1/2}}.$$
$$\frac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}=\frac{\left(a^{1/3}-b^{1/3}\right)\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}=a^{1/3}-b^{1/3}.$$
$$\frac{a^{0,5}-b^{0,5}}{a-b}=\frac{a^{1/2}-b^{1/2}}{\left(a^{1/2}-b^{1/2}\right)\left(a^{1/2}+b^{1/2}\right)}=\frac1{a^{1/2}+b^{1/2}}.$$
$$\frac{x^{3,5}y^{2,5}-x^{2,5}y^{3,5}}{x+2x^{0,5}y^{0,5}+y}=\frac{x^{2,5}y^{2,5}(x-y)}{\left(x^{0,5}+y^{0,5}\right)^2}.$$
Так как
$$x-y=\left(x^{0,5}-y^{0,5}\right)\left(x^{0,5}+y^{0,5}\right),$$
то
$$\frac{x^{2,5}y^{2,5}(x-y)}{\left(x^{0,5}+y^{0,5}\right)^2}=\frac{x^{2,5}y^{2,5}\left(x^{0,5}-y^{0,5}\right)}{x^{0,5}+y^{0,5}}.$$$$\frac{a-125}{a^{2/3}-25}=\frac{a^{1/3\,3}-5^3}{a^{2/3}-5^2}=\frac{\left(a^{1/3}-5\right)\left(a^{2/3}+5a^{1/3}+25\right)}{\left(a^{1/3}-5\right)\left(a^{1/3}+5\right)}=\frac{a^{2/3}+5a^{1/3}+25}{a^{1/3}+5}.$$
$$\frac{m^{7/6}-36m^{5/6}}{m^{1/2}-6m^{1/3}}=\frac{m^{5/6}\left(m^{1/3}-36\right)}{m^{1/3}\left(m^{1/6}-6\right)}.$$
Представим числитель как разность квадратов:
$$m^{1/3}-36=\left(m^{1/6}-6\right)\left(m^{1/6}+6\right).$$
Тогда
$$\frac{m^{5/6}\left(m^{1/6}-6\right)\left(m^{1/6}+6\right)}{m^{1/3}\left(m^{1/6}-6\right)}=m^{1/2}\left(m^{1/6}+6\right).$$$$\frac{24^{1/4}-8^{1/4}}{6^{1/4}-2^{1/4}}=\frac{2\cdot 6^{1/4}-2\cdot 2^{1/4}}{6^{1/4}-2^{1/4}}=2.$$
Ответ
- $$a^{1/3}$$
- $$\frac1n-\frac1m$$
- $$1+\frac{b}{a^{1/2}}$$
- $$a^{1/3}-b^{1/3}$$
- $$\frac1{a^{1/2}+b^{1/2}}$$
- $$\frac{x^{2,5}y^{2,5}\left(x^{0,5}-y^{0,5}\right)}{x^{0,5}+y^{0,5}}$$
- $$\frac{a^{2/3}+5a^{1/3}+25}{a^{1/3}+5}$$
- $$m^{1/2}\left(m^{1/6}+6\right)$$
- $$2$$