Wyznaczyć wszystkie rzeczywiste rozwiązania układu nierówności:
\(\displaystyle{ \begin{cases} 2xy-z ^{2} \ge 1 , \\ z-\left| x+y\right| \ge -1 . \end{cases} }\)
Układ nierówności
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RagaiH
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- Rejestracja: 16 lis 2023, o 19:29
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Re: Układ nierówności
\(\displaystyle{ \begin{cases} 2xy-z ^{2} \ge 1 &\text{(1) } \\ z-\left| x+y\right| \ge -1 &\text{(2) } \end{cases} }\)
(2):
\(\displaystyle{ z+1 \ge \left| x+y\right| }\)
\(\displaystyle{ (z+1) ^{2} \ge (x+y) ^{2} }\)
\(\displaystyle{ z ^{2} +2z+1 -x ^{2}-2xy-y ^{2} \ge 0 }\)
(1):
\(\displaystyle{ 2xy-z ^{2} \ge 1/ \cdot 2}\)
\(\displaystyle{ 4xy-2z ^{2}-2 \ge 0 }\)
(1)+(2):
\(\displaystyle{ z ^{2}-2z ^{2}+2z+1-2-x ^{2}-2xy+4xy-y ^{2} \ge 0 }\)
\(\displaystyle{ -z ^{2} +2z-1-x ^{2}+2xy-y ^{2} \ge 0/ \cdot (-1)}\)
\(\displaystyle{ z ^{2} -2z+1+x ^{2}-2xy+y ^{2} \le 0 }\)
\(\displaystyle{ (z-1) ^{2} +(x-y) ^{2} \le 0}\)
suma kwadratów nieujemna zatem:
\(\displaystyle{ z-1=0 \Rightarrow z=1}\)
\(\displaystyle{ x-y=0 \Rightarrow x=y}\)
(1):
\(\displaystyle{ 2xy-z ^{2} \ge 1}\)
\(\displaystyle{ 2x ^{2} \ge 2}\)
\(\displaystyle{ x ^{2} \ge 1 }\)
\(\displaystyle{ x \le -1 \vee x \ge 1 }\)
\(\displaystyle{ (2):}\)
\(\displaystyle{ z-\left| z+y\right| \ge -1}\)
\(\displaystyle{ \left| 2x\right| \le 2}\)
\(\displaystyle{ x \le 1 \wedge x \ge -1}\)
(1) \(\displaystyle{ \wedge}\) (2):
\(\displaystyle{ x=-1 \vee x=1}\)
zatem \(\displaystyle{ (x, y, z)=(1, 1, 1)}\) lub \(\displaystyle{ (-1, -1, 1)}\).
(2):
\(\displaystyle{ z+1 \ge \left| x+y\right| }\)
\(\displaystyle{ (z+1) ^{2} \ge (x+y) ^{2} }\)
\(\displaystyle{ z ^{2} +2z+1 -x ^{2}-2xy-y ^{2} \ge 0 }\)
(1):
\(\displaystyle{ 2xy-z ^{2} \ge 1/ \cdot 2}\)
\(\displaystyle{ 4xy-2z ^{2}-2 \ge 0 }\)
(1)+(2):
\(\displaystyle{ z ^{2}-2z ^{2}+2z+1-2-x ^{2}-2xy+4xy-y ^{2} \ge 0 }\)
\(\displaystyle{ -z ^{2} +2z-1-x ^{2}+2xy-y ^{2} \ge 0/ \cdot (-1)}\)
\(\displaystyle{ z ^{2} -2z+1+x ^{2}-2xy+y ^{2} \le 0 }\)
\(\displaystyle{ (z-1) ^{2} +(x-y) ^{2} \le 0}\)
suma kwadratów nieujemna zatem:
\(\displaystyle{ z-1=0 \Rightarrow z=1}\)
\(\displaystyle{ x-y=0 \Rightarrow x=y}\)
(1):
\(\displaystyle{ 2xy-z ^{2} \ge 1}\)
\(\displaystyle{ 2x ^{2} \ge 2}\)
\(\displaystyle{ x ^{2} \ge 1 }\)
\(\displaystyle{ x \le -1 \vee x \ge 1 }\)
\(\displaystyle{ (2):}\)
\(\displaystyle{ z-\left| z+y\right| \ge -1}\)
\(\displaystyle{ \left| 2x\right| \le 2}\)
\(\displaystyle{ x \le 1 \wedge x \ge -1}\)
(1) \(\displaystyle{ \wedge}\) (2):
\(\displaystyle{ x=-1 \vee x=1}\)
zatem \(\displaystyle{ (x, y, z)=(1, 1, 1)}\) lub \(\displaystyle{ (-1, -1, 1)}\).