Question

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

 

 

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

 

или можно выразить так:

 

 

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

 

или можно выразить так:

 

 

[tex]-frac{4sqrt{xy}}{sqrt{x}+sqrt{y}}" />