Упр.53 ГДЗ Колягин Ткачёва 9 класс (Алгебра)

1) ((a^18)^(1/3))^(1/3)+(v(a^4)^(1/3))^3;
2) (v(x^2)^(1/3))^3+2((vx)^(1/4))^8;
3) 2vv(a^4 b^8)-((v(a^3 b^6))^(1/3))^2;
4) (v(x^6 y^12))^(1/3)-((xy^2)^(1/5))^5;
5) (v(x^8 y^2)^(1/4))^4-((x^2 y^8)^(1/4))^2;
6) (((a a^(1/5))^(1/5))^5-a^(1/5)):(a^2)^(1/10).

1. $$\left(\sqrt[3]{\sqrt[3]{a^{18}}}\right)^3+\left(\sqrt{\sqrt[3]{a^4}}\right)^3$$
   $$=\sqrt[9]{a^{18}}+\sqrt{a^4}=a^2+a^2=2a^2.$$

2. $$\left(\sqrt{\sqrt[3]{x^2}}\right)^3+2\left(\sqrt[4]{\sqrt{x}}\right)^8$$
   $$=\sqrt{x^2}+2\left(\sqrt[8]{x}\right)^8=x+2x=3x.$$

3. $$2\sqrt{\sqrt{a^4b^8}}-\left(\sqrt[3]{\sqrt{a^3b^6}}\right)^2$$
   $$=2\sqrt[4]{a^4b^8}-\sqrt[3]{a^3b^6}=2ab^2-ab^2=ab^2.$$

4. $$\sqrt[3]{\sqrt{x^6y^{12}}}-\left(\sqrt[5]{xy^2}\right)^5$$
   $$=\sqrt[6]{x^6y^{12}}-xy^2=xy^2-xy^2=0.$$

5. $$\left(\sqrt{\sqrt[4]{x^8y^2}}\right)^4-\left(\sqrt[4]{x^2y^8}\right)^2$$
   $$=\sqrt{x^8y^2}-\sqrt{x^2y^8}=x^4y-xy^4.$$

6. $$\left(\left(\sqrt[5]{a\sqrt[5]{a}}\right)^5-\sqrt[5]{a}\right):\sqrt[10]{a^2}$$
   $$=\frac{a\sqrt[5]{a}-\sqrt[5]{a}}{\sqrt[5]{a}}=a-1.$$

Ответ

1) $$2a^2$$; 2) $$3x$$; 3) $$ab^2$$; 4) $$0$$; 5) $$x^4y-xy^4$$; 6) $$a-1$$.