Упр.23.18 ГДЗ Мерзляк Поляков 9 класс (Алгебра)
Сократите дробь:
- $$\frac{a+2a^{\frac{1}{3}}}{a^{\frac{2}{3}}+2}$$
- $$\frac{m^{\frac{5}{4}}n^{\frac{1}{4}}-m^{\frac{1}{4}}n^{\frac{5}{4}}}{m^{\frac{5}{4}}n^{\frac{5}{4}}}$$
- $$\frac{a-b^2}{a-a^{\frac{1}{2}}b}$$
- $$\frac{a-b}{a^{\frac{2}{3}}+a^{\frac{1}{3}}b^{\frac{1}{3}}+b^{\frac{2}{3}}}$$
- $$\frac{a^{0{,}5}-b^{0{,}5}}{a-b}$$
- $$\frac{x^{3{,}5}y^{2{,}5}-x^{2{,}5}y^{3{,}5}}{x+2x^{0{,}5}y^{0{,}5}+y}$$
- $$\frac{a-125}{a^{\frac{2}{3}}-25}$$
- $$\frac{m^{\frac{7}{6}}-36m^{\frac{5}{6}}}{m^{\frac{1}{2}}-6m^{\frac{1}{2}}}$$
- $$\frac{24^{\frac{1}{4}}-8^{\frac{1}{4}}}{6^{\frac{1}{4}}-2^{\frac{1}{4}}}$$
$$\frac{a+2a^{1/3}}{a^{2/3}+2}=\frac{a^{1/3}\left(a^{2/3}+2\right)}{a^{2/3}+2}=a^{1/3}.$$
$$\frac{m^{5/4}n^{1/4}-m^{1/4}n^{5/4}}{m^{5/4}n^{5/4}}= \frac{m^{1/4}n^{1/4}\left(m-n\right)}{m^{5/4}n^{5/4}}= \frac{m-n}{mn}=\frac1n-\frac1m.$$
$$\frac{a-b^2}{a-a^{1/2}b}= \frac{\left(a^{1/2}-b\right)\left(a^{1/2}+b\right)}{a^{1/2}\left(a^{1/2}-b\right)}= \frac{a^{1/2}+b}{a^{1/2}}=1+\frac{b}{a^{1/2}}.$$
$$\frac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}= \frac{\left(a^{1/3}-b^{1/3}\right)\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}= a^{1/3}-b^{1/3}.$$
$$\frac{a^{0,5}-b^{0,5}}{a-b}= \frac{\left(a^{1/2}-b^{1/2}\right)}{\left(a^{1/2}-b^{1/2}\right)\left(a^{1/2}+b^{1/2}\right)}= \frac1{a^{1/2}+b^{1/2}}.$$
$$\frac{x^{3,5}y^{2,5}-x^{2,5}y^{3,5}}{x+2x^{0,5}y^{0,5}+y}= \frac{x^{2,5}y^{2,5}(x-y)}{\left(x^{0,5}+y^{0,5}\right)^2}.$$
Так как $$x-y=\left(x^{0,5}-y^{0,5}\right)\left(x^{0,5}+y^{0,5}\right),$$ то
$$\frac{x^{3,5}y^{2,5}-x^{2,5}y^{3,5}}{x+2x^{0,5}y^{0,5}+y}= x^{2,5}y^{2,5}\cdot\frac{x^{0,5}-y^{0,5}}{x^{0,5}+y^{0,5}}.$$$$\frac{a-125}{a^{2/3}-25}= \frac{a^{1/3}-5}{a^{2/3}-5^2}= \frac{\left(a^{1/3}-5\right)\left(a^{2/3}+5a^{1/3}+25\right)}{\left(a^{1/3}-5\right)\left(a^{1/3}+5\right)}= \frac{a^{2/3}+5a^{1/3}+25}{a^{1/3}+5}.$$
$$\frac{m^{7/6}-36m^{5/6}}{m^{1/2}-6m^{1/2}}= \frac{m^{5/6}\left(m^{1/3}-36\right)}{m^{1/2}\left(1-6\right)}.$$
Упростим числитель:
$$m^{1/3}-36=\left(m^{1/6}-6\right)\left(m^{1/3}+6m^{1/6}+36\right).$$
Тогда
$$\frac{m^{7/6}-36m^{5/6}}{m^{1/2}-6m^{1/2}}= m^{1/3}\left(m^{1/6}+6\right).$$$$\frac{24^{1/4}-8^{1/4}}{6^{1/4}-2^{1/4}}= \frac{2\cdot 6^{1/4}-2\cdot 2^{1/4}}{6^{1/4}-2^{1/4}}= 2\cdot\frac{6^{1/4}-2^{1/4}}{6^{1/4}-2^{1/4}}=2.$$
Ответ
- $$a^{1/3}$$
- $$\frac1n-\frac1m$$
- $$1+\frac{b}{a^{1/2}}$$
- $$a^{1/3}-b^{1/3}$$
- $$\frac1{a^{1/2}+b^{1/2}}$$
- $$x^{2,5}y^{2,5}\cdot\frac{x^{0,5}-y^{0,5}}{x^{0,5}+y^{0,5}}$$
- $$\frac{a^{2/3}+5a^{1/3}+25}{a^{1/3}+5}$$
- $$m^{1/3}\left(m^{1/6}+6\right)$$
- $$2$$
