Uždavinys su laipsniais ir šaknim

$$\frac{14\sqrt{3}}{2\sqrt{3}+\sqrt{5}}$$

$$\frac{14\sqrt{3}}{2\sqrt{3}+\sqrt{5}}= \frac{14\sqrt{3}}{2\sqrt{3}+\sqrt{5}} \cdot \frac{2\sqrt{3}- \sqrt{5}}{2\sqrt{3}- \sqrt{5}}=\\\frac{14\sqrt{3} \cdot (2\sqrt{3}-\sqrt{5})}{12-5}=\\\frac{84-14\sqrt{15}}{7}=\frac{7(12-2\sqrt{15})}{7}=\\12-2\sqrt{15}=2(6-\sqrt{15})$$

Geriau nieko nesugalvoju :)

[tex]7^{\frac{1}{2}}\cdot 7^{\frac{1}{4}}\cdot 7^{\frac{1}{8}}\cdot 7^{\frac{1}{8}}=7[/tex]

$$\left ( 0,25 \right )^{-2} + 3^{-3}\ast \left ( \frac{1}9{} \right )^{-3} +\left ( 1,2 \right )^{0}$$

[tex](\frac{1}{4})^{-2}+3^{-3} \cdot (\frac{1}{9})^{-3}+(1.2)^{0}=4^{2}+3^{-3} \cdot 3^{6}+1=16+27+1=44[/tex]

$$\left ( \sqrt{5\sqrt{3}-\sqrt{11}}-\sqrt{5\sqrt{3} +\sqrt{11}} \right )^{2}$$

Naudosim šias formules: (x - y)² = x² - 2xy + y² ir (x - y)(x + y) = x² - y²

$$\left ( \sqrt{5\sqrt{3}-\sqrt{11}}-\sqrt{5\sqrt{3} +\sqrt{11}} \right )^{2}=\\ 5\sqrt{3}-\sqrt{11}-2\sqrt{(5\sqrt{3}-\sqrt{11})(5\sqrt{3}+\sqrt{11})}+5\sqrt{3}+\sqrt{11}=\\10\sqrt{3}-2\sqrt{75-11}=10\sqrt{3}-16$$