Упростите выражение.

1) $\frac{3}{x^\frac{1}{4}+y^\frac{1}{4}} +\frac{3}{x^\frac{1}{4}-y^\frac{1}{4}} =\frac{3x^\frac{1}{4}-3y^\frac{1}{4}+3x^\frac{1}{4}+3y^\frac{1}{4}}{(x^\frac{1}{4}+y^\frac{1}{4})*(x^\frac{1}{4}-y^\frac{1}{4})} =\frac{6x^\frac{1}{4} }{x^\frac{1}{2}-y^\frac{1}{2}}$

2) $\frac{6x^\frac{1}{4} }{x^\frac{1}{2}-y^\frac{1}{2}}*\frac{(y^{-\frac{1}{2}} -x^{-\frac{1}{2}} )}{4*x^{-\frac{1}{4}}*y^{-\frac{1}{2}}} =\frac{6*x^{\frac{1}{4}}*x^{\frac{1}{4}}*y^{\frac{1}{2}}*(y^{-\frac{1}{2}} -x^{-\frac{1}{2}} ) }{(x^\frac{1}{2}-y^\frac{1}{2})*4 } =\frac{6*x^{\frac{1}{2}}*y^{\frac{1}{2}}*(y^{-\frac{1}{2}} -x^{-\frac{1}{2}} ) }{(x^\frac{1}{2}-y^\frac{1}{2})*4 }=\frac{6*x^{\frac{1}{2}}-6*y^{\frac{1}{2}} }{(x^\frac{1}{2}-y^\frac{1}{2})*4 } =\frac{6}{4}=1.5$

Ответ:1.5