Да се опрости

[Гост]

н

[Евва]

За удобство полагаме: [tex]\sqrt{a-b}[/tex]=x ,[tex]\sqrt{a+b}[/tex]=y
[tex]y^{2}[/tex]=a+b ,[tex]x^{2}[/tex]=a-b [tex]\Rightarrow[/tex] [tex]y^{2}[/tex]-[tex]x^{2}[/tex]=2b [tex]\Rightarrow[/tex] b=[tex]\frac{y^{2}-x^{2}}{2}[/tex]

Задачата придобива следния вид:
( [tex]\frac{х}{у+х}[/tex]+[tex]\frac{x^{2}}{ху-x^{2}}[/tex] )[tex]\frac{\frac{у^{2}-x^{2}}{2}}{ху}[/tex]=

=( [tex]\frac{х}{у+х}[/tex]+[tex]\frac{x^{2}}{х(у-х)}[/tex] )[tex]\frac{(у-х)(у+х)}{2ху}[/tex]=

=( [tex]\frac{х}{у+х}[/tex]+[tex]\frac{х}{у-х}[/tex] )[tex]\frac{(у-х)(у+х)}{2ху}[/tex]=

=[tex]\frac{ху-x^{2}+ху+x^{2}}{(у-х)(у+х)}[/tex] .[tex]\frac{(у-х)(у+х)}{2ху}[/tex]=

=1

[S.B.]

Още един поглед върху задачата

[tex](\frac{\sqrt{a - b}}{\sqrt{a + b} + \sqrt{a - b}} + \frac{a - b}{\sqrt{a^{2} - b^{2}} - a + b}).\frac{b}{\sqrt{a^{2} - b^{2}}} =[/tex]

$=[\frac{(a - b)^{\frac{1}{2}}}{(a + b)^{\frac{1}{2}} + (a - b)^{\frac{1}{2}}} + \frac{(a - b)}{(a - b)^{\frac{1}{2}}.(a + b)^{\frac{1}{2}} - (a - b)}].\frac{b}{(a - b)^{\frac{1}{2}}.(a + b)^{\frac{1}{2}}} = $

$= [\frac{(a - b)^{\frac{1}{2}}}{(a + b)^{\frac{1}{2}} + (a - b)^{\frac{1}{2}}} + \frac{(a - b)}{(a - b)^{\frac{1}{2}}((a + b)^{\frac{1}{2}} -(a - b)^{\frac{1}{2}})}].\frac{b}{(a + b)^{\frac{1}{2}}(a - b)^{\frac{1}{2}}} = $

$=[\frac{(a - b)^{\frac{1}{2}}}{(a + b)^{\frac{1}{2}} + (a - b)^{\frac{1}{2}}} + \frac{(a - b)^{\frac{1}{2}}}{(a + b)^{\frac{1}{2}} - (a - b)^{\frac{1}{2}} }] . \frac{b}{(a + b)^{\frac{1}{2}}.(a - b)^{\frac{1}{2}}} = $

$ = [\frac{1}{(a + b)^{\frac{1}{2}} + (a - b)^{\frac{1}{2}}} + \frac{1}{(a + b)^{\frac{1}{2}} - (a - b)^{\frac{1}{2}}}].\frac{b(a - b)^{\frac{1}{2}}}{(a + b)^{\frac{1}{2}}(a - b)^{\frac{1}{2}}} = $

$ = \frac{(a + b)^{\frac{1}{2}} - (a - b)^{\frac{1}{2}} + (a + b)^{\frac{1}{2}} + (a - b)^{\frac{1}{2}}}{a + b - a + b}.\frac{b}{(a + b)^{\frac{1}{2}}} = $
$ = \frac{2(a + b)^{\frac{1}{2}}}{2b}.\frac{b}{(a + b)^{\frac{1}{2}}} = $

$ = \frac{2}{2} = 1$