Estude! Com Prof. Caju

{[tex3]2^{\sqrt{x}+\sqrt{y}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]2^{\sqrt{x}+\sqrt{y}}[/tex3]

= 512
{log [tex3]\sqrt{xy}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sqrt{xy}[/tex3]

= 1+ log 2

(IEZZI) Resolva o sistema de equações acima:

Gab: S: (25 , 16) , (16, 25)

[tex3]x > 0, y > 0\\
2^{\sqrt{x}+\sqrt{y}}=2^9 \implies{\sqrt x}+{\sqrt y} = 9(I)\\
log{\sqrt{ (xy)}}-log2=1 \implies
log({\frac{\sqrt{xy}}{2}})=log10 \implies \sqrt{xy} = 20 \implies xy = 400 (II)\\
(\sqrt x+\sqrt y)^2 = 9 \implies x+y+2\sqrt{xy} = 81 \implies x+y =41 \implies x = 41-y(III)\\
(III)em(II): (41-y). y = 400 \implies y^2-41y+400 = 0 \\
\therefore y = 16 \vee y = 25 \implies x = 25 \vee x = 16

[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]x > 0, y > 0\\
2^{\sqrt{x}+\sqrt{y}}=2^9 \implies{\sqrt x}+{\sqrt y} = 9(I)\\
log{\sqrt{ (xy)}}-log2=1 \implies
log({\frac{\sqrt{xy}}{2}})=log10 \implies \sqrt{xy} = 20 \implies xy = 400 (II)\\
(\sqrt x+\sqrt y)^2 = 9 \implies x+y+2\sqrt{xy} = 81 \implies x+y =41 \implies x = 41-y(III)\\
(III)em(II): (41-y). y = 400 \implies y^2-41y+400 = 0 \\
\therefore y = 16 \vee y = 25 \implies x = 25 \vee x = 16

[/tex3]

Muito Obrigado Petras!