Упр.10.24 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

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

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

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

5) $$\frac{a^{0.5}-b^{0.5}}{a-b}=\frac{a^{1/2}-b^{1/2}}{(a^{1/2}-b^{1/2})(a^{1/2}+b^{1/2})}=\frac1{a^{1/2}+b^{1/2}}.$$

6) $$\frac{x^{3.5}y^{2.5}-x^{2.5}y^{3.5}}{x+2x^{0.5}y^{0.5}+y}=\frac{x^{2.5}y^{2.5}(x-y)}{(x^{0.5}+y^{0.5})^2}=\frac{x^{2.5}y^{2.5}(x^{0.5}-y^{0.5})(x^{0.5}+y^{0.5})}{(x^{0.5}+y^{0.5})^2}$$

$$=\frac{x^{2.5}y^{2.5}(x^{0.5}-y^{0.5})}{x^{0.5}+y^{0.5}}.$$

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

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

$$=\frac{m^{5/6}(m^{1/6}-6)(m^{1/6}+6)}{m^{1/3}(m^{1/6}-6)}=m^{1/2}(m^{1/6}+6).$$

9) $$\frac{24^{1/4}-8^{1/4}}{6^{1/4}-2^{1/4}}=\frac{2^{1/4}(3^{1/4}-1)}{2^{1/4}(3^{1/4}-1)}=2^{1/2}.$$

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) $$\frac{x^{2.5}y^{2.5}(x^{0.5}-y^{0.5})}{x^{0.5}+y^{0.5}}$$; 7) $$\frac{a^{2/3}+5a^{1/3}+25}{a^{1/3}+5}$$; 8) $$m^{1/2}(m^{1/6}+6)$$; 9) $$2^{1/2}$$.