Question

Решите пожалуйста систему уравнений

Answer 1

Ответ: $a=-2\sqrt{6},b=-2\sqrt{2}\\a=-2\sqrt{6},b=-\frac{2\sqrt{2}}{5}\\a=2\sqrt{6},b=2\sqrt{2}\\a=2\sqrt{6},b=\frac{2\sqrt{2}}{5}$

Пошаговое объяснение:

$\left\{{{\frac{b^2\sqrt{3}}{4}+\frac{a^2\sqrt{3}}{2}+\frac{(\sqrt{3}b-a)^2\sqrt{3}}{8}=14\sqrt{3}}\atop{\frac{b^2\sqrt{3}}{4}+\frac{(\sqrt{3}b-a)^2\sqrt{3}}{8}=2\sqrt{3}}}\right. \\\left\{{{\frac{b^2}{4}+\frac{a^2}{2}+\frac{(\sqrt{3}b-a)^2}{8}=14}\atop{\frac{b^2}{4}+\frac{(\sqrt{3}b-a)^2}{8}=2}}\right.\\\left\{{{\frac{a^2}{2}=12}\atop{\frac{b^2}{4}+\frac{(\sqrt{3}b-a)^2}{8}=2}}\right.\\\left\{{{a=-2\sqrt{6},a=2\sqrt{6}}\atop{\frac{b^2}{4}+\frac{(\sqrt{3}b-a)^2}{8}=2}}\right.$

$\left\{{{a=-2\sqrt{6};a=2\sqrt{6}}\atop{\frac{b^2}{4}+\frac{(\sqrt{3}b+2\sqrt{6})^2}{8}=2,\frac{b^2}{4}+\frac{(\sqrt{3}b-2\sqrt{6})^2}{8}=2}}\right.\\\left\{{{a=-2\sqrt{6};a=2\sqrt{6}}\atop{\frac{b^2}{4}+\frac{(\sqrt{3}b+2\sqrt{6})^2}{8}=2,\frac{b^2}{4}+\frac{(\sqrt{3}b-2\sqrt{6})^2}{8}=2}}\right.\\\left\{{{a=-2\sqrt{6};a=2\sqrt{6}}\atop{5b^2+12\sqrt{2}b+8;5b^2-12\sqrt{2}b+8}}\right.$

$\left\{{{a=-2\sqrt{6};a=2\sqrt{6}}\atop{(b=-2\sqrt{2};b=-\frac{2\sqrt{2}}{5});(b=2\sqrt{2};b=\frac{2\sqrt{2}}{5})}}\right.$