Упр.394 ГДЗ Никольский Потапов 9 класс (Алгебра)

9. а) $$\left(2x^{\frac{1}{2}}-y^{\frac{1}{3}}\right)\left(y^{\frac{1}{3}}+2x^{\frac{1}{2}}\right)$$
   б) $$\left(3m^{\frac{1}{4}}-n^{\frac{1}{4}}\right)\left(n^{\frac{1}{4}}+3m^{\frac{1}{4}}\right)$$
   в) $$\left(x^{\frac{1}{6}}-5\right)\left(x^{\frac{1}{6}}+5\right)$$
   г) $$\left(3^{\frac{1}{5}}-x^{\frac{2}{3}}\right)\left(3^{\frac{1}{5}}+x^{\frac{2}{3}}\right)$$

Используем формулу разности квадратов:

$$(a-b)(a+b)=a^2-b^2.$$

1. $$(2x^{1/2}-y^{1/3})(y^{1/3}+2x^{1/2}) =(2x^{1/2})^2-(y^{1/3})^2 =4x-y^{2/3}.$$

2. $$(3m^{1/4}-n^{1/4})(n^{1/4}+3m^{1/4}) =(3m^{1/4})^2-(n^{1/4})^2 =9m^{1/2}-n^{1/2}.$$

3. $$(x^{1/6}-5)(x^{1/6}+5) =(x^{1/6})^2-5^2 =x^{1/3}-25.$$

4. $$(3^{1/5}-x^{2/3})(3^{1/5}+x^{2/3}) =(3^{1/5})^2-(x^{2/3})^2 =3^{2/5}-x^{4/3}.$$

Ответ

а) $$4x-y^{2/3}$$; б) $$9m^{1/2}-n^{1/2}$$; в) $$x^{1/3}-25$$; г) $$3^{2/5}-x^{4/3}$$.