Question

Помогите пожалуйста ​

Answer 1

$(\sqrt{27}-\sqrt{8}+\sqrt{32}-\sqrt{12})^{-2}=\\(3\sqrt{3}-2\sqrt{2}+4\sqrt{2}-2\sqrt{3})^{-2}=\\(\sqrt{3}+2\sqrt{2})^{-2}=\frac{1}{3+8+4\sqrt{6}}^{(11-4\sqrt{6}}=\\\frac{11-4\sqrt{6}}{121-96}=\frac{11-4\sqrt{6}}{25}\\Otvet:\frac{11-4\sqrt{6}}{25}.$

$\frac{11-4\sqrt{6}}{25}=\frac{4(11-4\sqrt{6})}{100};\sqrt{6}\approx2.45\\\frac{11-4\sqrt{6}}{25}\approx \frac{4(11-9.8)}{100}=\frac{4*1.2}{100}=0.048$