Question

Решите пожалуйста систему уравнений

Answer 1

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

Пошаговое объяснение:

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

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

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