Упр.37.19 ГДЗ Мордкович 10-11 класс (Алгебра)

а) (a^0,4)^1/2 * a^0,8;
б) (10)корень(c) * (c^-1,2)^3/4;
в) (x^3/4)^5/4 * ((4)корень(x))^17/4;
г) (b^0,8)^-3/4 * (b^-2/5)^-1,5.

1. $$\left(a^{0,4}\right)^{\frac12}\cdot a^{0,8}=\left(a^{\frac25}\right)^{\frac12}\cdot a^{\frac45}=a^{\frac25\cdot\frac12}\cdot a^{\frac45}=a^{\frac15+\frac45}=a^1=a.$$
2. $$\sqrt[10]{c}\cdot\left(c^{-1,2}\right)^{\frac34}=c^{\frac1{10}}\cdot\left(c^{-\frac65}\right)^{\frac34}=c^{\frac1{10}}\cdot c^{-\frac65\cdot\frac34}=c^{\frac1{10}-\frac9{10}}=c^{-\frac45}=\frac{1}{\sqrt[5]{c^4}}.$$
3. $$\left(x^{\frac34}\right)^{\frac54}\cdot\left(\sqrt[4]{x}\right)^{\frac{17}{4}}=\left(x^{\frac34}\right)^{\frac54}\cdot\left(x^{\frac14}\right)^{\frac{17}{4}}=x^{\frac34\cdot\frac54}\cdot x^{\frac14\cdot\frac{17}{4}}=x^{\frac{15}{16}+\frac{17}{16}}=x^2.$$
4. $$\left(b^{0,8}\right)^{-\frac34}\cdot\left(b^{-\frac25}\right)^{-1,5}=\left(b^{\frac45}\right)^{-\frac34}\cdot\left(b^{-\frac25}\right)^{-\frac32}=b^{\frac45\cdot\left(-\frac34\right)}\cdot b^{-\frac25\cdot\left(-\frac32\right)}=b^{-\,\frac35}\cdot b^{\frac35}=b^0=1.$$

Ответ

а) $$a$$; б) $$\frac{1}{\sqrt[5]{c^4}}$$; в) $$x^2$$; г) $$1$$.