Question

5+y^0.5)/(x^0.5-y^0..5-y^0.5)/(x^0.5+y^0.*(y^0.5-x^0.5)

Answer 1

((y/x)^0.5 -(x/y)^0.5)-(x/y)^0.5+(y/x)^0.5)*(y^0.5-x^0.5)=

=2((y/x)^0.5 -(x/y)^0.5)*(y^0.5-x^0.5)=

=2(y/x^0,5+y^0.5-y^0.5+x/y^0.5)=

=2(y/x^0.5+x/y^0.5)

Answer 2

$(\frac{x^{0,5}+y^{0,5}}{x^{0,5}-y^{0,5}}-\frac{x^{0,5}-y^{0,5}}{x^{0,5}+y^{0,5}})(y^{0,5}-x^{0,5})$

произведём замену переменных:

$a=x^{0,5}$$b=y^{0,5}$

$(\frac{a+b}{a-b}-\frac{a-b}{a+b})(b-a)= \\ \\=(\frac{a+b}{a-b}-\frac{a-b}{a+b})(-a+b)= \\ \\=-(\frac{a+b}{a-b}-\frac{a-b}{a+b})(a-b)= \\ \\=-(\frac{(a+b)^{2}}{(a-b)(a+b)}-\frac{(a-b)^{2}}{(a+b)(a-b)})(a-b)= \\ \\=-(\frac{(a^{2}+2ab+b^{2})-(a^{2}-2ab+b^{2})}{(a-b)(a+b)})(a-b)= \\ \\=-\frac{a^{2}+2ab+b^{2}-a^{2}+2ab-b^{2}}{(a-b)(a+b)}(a-b)= \\ \\=-\frac{4ab(a-b)}{(a-b)(a+b)}= \\ \\=-\frac{4(ab)}{(a+b)}$

произведём обратную замену переменных:

$-\frac{4(xy)^{0,5}}{(x^{0,5}+y^{0,5})}$

или можно выразить так:

$-\frac{4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}$