Question

помогите 55 балов даю срочно

Answer 1

Ответ:

$-1,375$

Объяснение:

$\displaystyle\\\frac{3y+x}{3y-x}+\frac{3y-x}{3y+x}=\frac{41}{20} \\\\\frac{(3y+x)^2+(3y-x)^2}{(3y+x)(3y-x)}=\frac{41}{20} \\\\\frac{9y^2+6xy+x^2+9y^2-6xy+x^2}{(3y)^2-x^2}=\frac{41}{20} \\\\ \frac{18y^2+2x^2}{9y^2-x^2}=\frac{41}{20} \\\\ 20(18y^2+2x^2)=41(9y^2-x^2)\\\\360y^2+40x^2=369y^2-41x^2\\\\81x^2=9y^2\\\\y^2=9x^2\\\\\frac{2x^2+y^2}{x^2-y^2} =\frac{2x^2+9x^2}{x^2-9x^2} =\frac{11x^2}{-8x^2}=-\frac{11}{8} =\boldsymbol{-1,375}$