1.107. Өрнекті ықшамдаңдар: 1) (0,25a⁻¹y⁻³)² · (a⁻³/(4y²))⁻³; 2) ((x⁻³y⁴)/9)⁻² · (3/(x⁻²y³))⁻³; 3) (m⁻⁴/(10n⁵k²))⁻² : (5m²n³k)³; 4) (9c⁵/(d³b⁻²))⁻² : (a²b⁻³/(6c⁴))³.

1. 0,25 = 1/4 болғандықтан:

   \(\left(\frac{a^{-1}}{4y^3}\right)^2\cdot\left(\frac{a^{-3}}{4y^2}\right)^{-3}=\frac{a^{-2}}{16y^6}\cdot\frac{64a^9y^6}{1}=4a^7.\)

2. Теріс дәреже бөлшекті кері айналдырады:

   \(\left(\frac{x^{-3}y^4}{9}\right)^{-2}\cdot\left(\frac{3}{x^{-2}y^3}\right)^{-3}=\frac{81x^6}{y^8}\cdot\frac{y^9}{27x^6}=3y.\)

3. \(\left(\frac{m^{-4}}{10n^5k^2}\right)^{-2}=100m^8n^{10}k^4\), ал \((5m^2n^3k)^3=125m^6n^9k^3\). Сондықтан:

   \(100m^8n^{10}k^4:125m^6n^9k^3=\frac45m^2nk.\)

4. \(\left(\frac{9c^5}{d^3b^{-2}}\right)^{-2}=\frac{d^6}{81b^4c^{10}}\), ал \(\left(\frac{a^2b^{-3}}{6c^4}\right)^3=\frac{a^6}{216b^9c^{12}}\). Демек:

   \(\frac{d^6}{81b^4c^{10}}:\frac{a^6}{216b^9c^{12}}=\frac{8b^5c^2d^6}{3a^6}.\)

Жауабы: 1) 4a⁷; 2) 3y; 3) ⁴⁄₅m²nk; 4) 8b⁵c²d⁶/(3a⁶).