Odpowiedź:
x² = - √2 y
y² = 0,5 x / * 2 ⇒ x = 2 y²
więc
( 2 y²)² = - √2 y
4 [tex]y^4 + \sqrt{2} y = 0[/tex]
√2 y*( 2√2 y³ + 1) = 0
2√2 y³ = - 1
y³ = - [tex]\frac{1}{2\sqrt{2} }[/tex]
y = - [tex]\frac{1}{\sqrt{2} }[/tex] ⇒ x = 2*( - [tex]\frac{1}{\sqrt{2} } )^2 = 2*\frac{1}{2} = 1[/tex]
zatem
[tex]\frac{1}{x + y} = \frac{1}{1 - \frac{1}{\sqrt{2} } } = \frac{\sqrt{2} }{\sqrt{2} - 1} *[/tex] [tex]\frac{\sqrt{2} + 1}{\sqrt{2} + 1 } =[/tex] [tex]\frac{\sqrt{2} *(\sqrt{2} +1)}{2 - 1} = 2 + \sqrt{2}[/tex]
===========================================
y = 0 i x = 0 = odpada
Szczegółowe wyjaśnienie:
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
Odpowiedź:
x² = - √2 y
y² = 0,5 x / * 2 ⇒ x = 2 y²
więc
( 2 y²)² = - √2 y
4 [tex]y^4 + \sqrt{2} y = 0[/tex]
√2 y*( 2√2 y³ + 1) = 0
2√2 y³ = - 1
y³ = - [tex]\frac{1}{2\sqrt{2} }[/tex]
y = - [tex]\frac{1}{\sqrt{2} }[/tex] ⇒ x = 2*( - [tex]\frac{1}{\sqrt{2} } )^2 = 2*\frac{1}{2} = 1[/tex]
zatem
[tex]\frac{1}{x + y} = \frac{1}{1 - \frac{1}{\sqrt{2} } } = \frac{\sqrt{2} }{\sqrt{2} - 1} *[/tex] [tex]\frac{\sqrt{2} + 1}{\sqrt{2} + 1 } =[/tex] [tex]\frac{\sqrt{2} *(\sqrt{2} +1)}{2 - 1} = 2 + \sqrt{2}[/tex]
===========================================
y = 0 i x = 0 = odpada
Szczegółowe wyjaśnienie: