DK; x≥y≥0pt(1) ⇔(1−y)(√x−y−1)+x−y−1−(x−y−1)√y=0 ⇔(x−y−1)(1−y√x−y+1+1−√y)=0 ⇔(x−y−1)(1−√y)(1+√y√x−y+1+1)=0 ⇔x−y−1=0 of y=1 do (...)>04+)y=1 \Rightarrow x=3 \Rightarrow (x;y)=(3;1)+)x=y+1 (2)\Leftrightarrow 2y^{2}+3y-2=\sqrt{1-y}\Leftrightarrow 2(y^{2}+y-1)+y-\sqrt{1-y}=04 ⇔(y2+y−1)(2+1y+√1−y)=0 ⇔y2+y−1=0 do (... )>0 ⇔y=−1+√52(tm) of y=−1−√52(L) ⇔(x;y)=(1+√52;−1+√52)
DK;
x≥y≥0pt(1)
⇔(1−y)(√x−y−1)+x−y−1−(x−y−1)√y=0 ⇔(x−y−1)(1−y√x−y+1+1−√y)=0 ⇔(x−y−1)(1−√y)(1+√y√x−y+1+1)=0 ⇔x−y−1=0 of y=1 do
(...)>04+)y=1 \Rightarrow x=3 \Rightarrow (x;y)=(3;1)
+)x=y+1
(2)\Leftrightarrow 2y^{2}+3y-2=\sqrt{1-y}
\Leftrightarrow 2(y^{2}+y-1)+y-\sqrt{1-y}=0
$ ⇔(y2+y−1)(2+1y+√1−y)=0 ⇔y2+y−1=0 do (... )>0
⇔y=−1+√52(tm) of
y=−1−√52(L) ⇔(x;y)=(1+√52;−1+√52)