Ответ:
[tex]0 \ ; \ 2 \ ;[/tex]
[tex](-4; \ -i\sqrt{11}) \ , \ (-4; \ i\sqrt{11}) \ , \ (-2; \ -1) \ , \ (2; \ -1) \ ;[/tex]
Объяснение:
[tex]11. \ 2^{x}+4 \cdot 2^{-x}=5 \quad | \quad \cdot 2^{x}[/tex]
[tex](2^{x})^{2}+4=5 \cdot 2^{x};[/tex]
[tex](2^{x})^{2}-5 \cdot 2^{x}+4=0;[/tex]
[tex](2^{x})^{2}-2^{x}-4 \cdot 2^{x}+4=0;[/tex]
[tex]2^{x} \cdot (2^{x}-1)-4 \cdot (2^{x}-1)=0;[/tex]
[tex](2^{x}-1) \cdot (2^{x}-4)=0;[/tex]
[tex]2^{x}-1=0 \quad \vee \quad 2^{x}-4=0;[/tex]
[tex]2^{x}=1 \quad \vee \quad 2^{x}=4;[/tex]
[tex]2^{x}=2^{0} \quad \vee \quad 2^{x}=2^{2};[/tex]
[tex]x=0 \quad \vee \quad x=2;[/tex]
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[tex]$ \displaystyle \left \{ {{x(1+y)+4=-4y} \atop {x^{2}+y^{2}=5}} \right. \Leftrightarrow \left \{ {{x(1+y)+4+4y=0} \atop {x^{2}+y^{2}=5}} \right. \Leftrightarrow \left \{ {{x(1+y)+4(1+y)=0} \atop {x^{2}+y^{2}=5}} \right. \Leftrightarrow $[/tex]
[tex]$ \displaystyle \Leftrightarrow \left \{ {{(x+4)(1+y)=0} \atop {x^{2}+y^{2}=5}} \right. \Leftrightarrow \left \{ {{x+4=0} \atop {x^{2}+y^{2}=5}} \right. \vee \left \{ {{1+y=0} \atop {x^{2}+y^{2}=5}} \right. \Leftrightarrow \left \{ {{x=-4} \atop {(-4)^{2}+y^{2}=5}} \right. \vee $[/tex]
[tex]$ \displaystyle \vee \left \{ {{y=-1} \atop {x^{2}+(-1)^{2}=5}} \right. \Leftrightarrow \left \{ {{x=-4} \atop {16+y^{2}=5}} \right. \vee \left \{ {{y=-1} \atop {x^{2}+1=5}} \right. \Leftrightarrow \left \{ {{x=-4} \atop {y^{2}=5-16}} \right. \vee $[/tex]
[tex]$ \displaystyle \vee \left \{ {{y=-1} \atop {x^{2}=5-1}} \right. \Leftrightarrow \left \{ {{x=-4} \atop {y^{2}=-11}} \right. \vee \left \{ {{y=-1} \atop {x^{2}=4}} \right. \Leftrightarrow \left \{ {{x=-4} \atop {y=\pm \sqrt{-11}}} \right. \vee \left \{ {{y=-1} \atop {x=\pm \sqrt{4}}} \right. \Leftrightarrow $[/tex]
[tex]$ \displaystyle \Leftrightarrow \left \{ {{x=-4} \atop {y=\pm i\sqrt{11}}} \right. \vee \left \{ {{x=\pm 2} \atop {y=-1}} \right. \ ;[/tex]
[tex](-4; \ -i\sqrt{11}) \ , \ (-4; \ i\sqrt{11}) \ , \ (-2; \ -1) \ , \ (2; \ -1) \ ;[/tex]
