Упр.23.11 ГДЗ Мерзляк Поляков 9 класс (Алгебра)

1) 2a^(1/2) (a^(1/2)-4)+8a^(1/2); 4) (a^(1/3)+b^(1/3))^2;
2) (a^0,5-3b^0,3)(2a^0,5+b^0,3); 5) (a^(-1/2)-(1/4)a^(-1/6))^2;
3) (3b^(2/3)-c^(3/2))(3b^(2/3)+c^(3/2); 6) (b^0,4+3)^2-6b^0,4;
7) (c^(1/3)-1)(c^(2/3)+c^(1/3)+1); 8) (a^(1/3)+a^(1/2))(a^(2/3)-a^(5/6)+a);
9) (a^(1/6)+b^(1/6))(a^(1/6)-b^(1/6))(a^(1/3)-b^(1/3));
10) (x^(2/9)-1)(x^(4/9)+x^(2/9)+1)(x^(2/3)+1).

1. $$2a^{1/2}(a^{1/2}-4)+8a^{1/2}=2a^{1/2}\cdot a^{1/2}-8a^{1/2}+8a^{1/2}=2a.$$
2. $$ (a^{0,5}-3b^{0,3})(2a^{0,5}+b^{0,3})= 2a^{1}+a^{0,5}b^{0,3}-6a^{0,5}b^{0,3}-3b^{0,6} $$
   $$ =2a-5a^{0,5}b^{0,3}-3b^{0,6} =2a-5\sqrt{a}\,\sqrt[10]{b^3}-3\sqrt[5]{b^3}. $$
3. $$ (3b^{2/3}-c^{3/2})(3b^{2/3}+c^{3/2})=9b^{4/3}-c^3=9b\sqrt[3]{b}-c^3. $$
4. $$ (a^{1/3}+b^{1/3})^2=a^{2/3}+2a^{1/3}b^{1/3}+b^{2/3} =\sqrt[3]{a^2}+2\sqrt[3]{ab}+\sqrt[3]{b^2}. $$
5. $$ \left(a^{-1/2}-\frac14 a^{-1/6}\right)^2 =a^{-1}-\frac12 a^{-2/3}+\frac1{16}a^{-1/3} =\frac1a-\frac1{2\sqrt[3]{a^2}}+\frac1{16\sqrt[3]{a}}. $$
6. $$ (b^{0,4}+3)^2-6b^{0,4}=b^{0,8}+9=b^{4/5}+9=\sqrt[5]{b^4}+9. $$
7. $$ (c^{1/3}-1)(c^{2/3}+c^{1/3}+1)=c-1. $$
8. $$ (a^{1/3}+a^{1/2})(a^{2/3}-a^{5/6}+a)=a+a^{1/2}=a+\sqrt{a}. $$
9. $$ (a^{1/6}+b^{1/6})(a^{1/6}-b^{1/6})(a^{1/3}-b^{1/3}) =(a^{1/3}-b^{1/3})(a^{1/3}-b^{1/3}) $$
   $$ =a^{2/3}-2a^{1/3}b^{1/3}+b^{2/3} =\sqrt[3]{a^2}-2\sqrt[3]{ab}+\sqrt[3]{b^2}. $$
10. $$ (x^{2/9}-1)(x^{4/9}+x^{2/9}+1)(x^{2/3}+1) =(x^{2/3}-1)(x^{2/3}+1) $$
    $$ =x^{4/3}-1=\sqrt[3]{x^4}-1. $$

Ответ

1) $$2a$$; 2) $$2a-5\sqrt{a}\,\sqrt[10]{b^3}-3\sqrt[5]{b^3}$$; 3) $$9b\sqrt[3]{b}-c^3$$; 4) $$\sqrt[3]{a^2}+2\sqrt[3]{ab}+\sqrt[3]{b^2}$$; 5) $$\frac1a-\frac1{2\sqrt[3]{a^2}}+\frac1{16\sqrt[3]{a}}$$; 6) $$\sqrt[5]{b^4}+9$$; 7) $$c-1$$; 8) $$a+\sqrt{a}$$; 9) $$\sqrt[3]{a^2}-2\sqrt[3]{ab}+\sqrt[3]{b^2}$$; 10) $$\sqrt[3]{x^4}-1$$.