Помогите решить 6.1 6.2

Решите задачу:

$6.1)\; \; \Big (\frac{5a^3b^5}{3c^7d^2}\Big )^4=\frac{625\; a^{12}\; b^{20}}{81\; c^{28}\; d^8}\\\\\Big (\frac{2x^{-3}y^2}{3x^4y^{-5}}\Big )^{-2}=\Big (\frac{2y^7}{3x^7}\Big )^{-2}=\Big (\frac{3x^7}{2y^7}\Big )^2=\frac{9\; x^{14}}{4\; y^{14}}=\frac{9}{4}\cdot \Big (\frac{x}{y}\Big )^{14}$

$\Big (\frac{5\, u^{-4}v^3\, t^{-5}}{2^{-2}\, u^{-3}\, v^{-1}\, t^3}\Big )^{-3/2}=\Big (\frac{5\, u^{-1}\, v^{4}\, t^{-8}}{\frac{1}{4}}\Big )^{-3/2}=(20)^{-3/2}\, u^{3/2}\, v^{-6}\, t^{12}=\frac{u^{3/2}\, t^{12}}{20^{3/2}\, v^6}\\\\\Big (\frac{x^{-1/2}\, y^{2/3}\, z^2}{x^2\, y^{-1/3}\, z^{1/3}}\Big )^{1/2}=\Big (\frac{y\cdot z^{5/3}}{x^{5/2}}\Big )^{1/2}=\frac{y^{1/2}\, z^{5/6}}{x^{5/4}}\\\\6.2)\; \; \sqrt9\cdot \sqrt{16}-\sqrt{25}\cdot \sqrt{36}=3\cdot 4-5\cdot 6=12-30=-18\\\\\sqrt{75^2-2\cdot 75\cdot 74+74^2}=\sqrt{(75-74)^2}=\sqrt{1^2}=1}$