система уравнений решите 1)x^2+y^2=20 xy=8 2)x^2+y^2=5 xy=2. Created by Dogg11. algebra-ru

Пусть t = y². t₁ + t₂ = 20t₁*t₂ = 64t₁ = 4t₂ = 16Обратная замена:y² = 4y² = 16Ответ: 2) Пусть z = y².z² - 5z + 4 = 0z₁ + z₂ = 5z₁*z₂ = 4z₁ = 4z₂ = 1Обратная замена:Ответ:

система уравнений решите 1)x^2+y^2=20 xy=8 2)x^2+y^2=5 xy=2...

Dogg11 @Dogg11

July 2022 1 0 Report

система уравнений решите
1)x^2+y^2=20
xy=8

2)x^2+y^2=5
xy=2

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Dимасuk

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$\left \{ {{ x^{2} +y^{2} = 20} \atop {xy=8}} \right. 
$
$\left \{ {{ x^{2} +y^{2} = 20} \atop {x= \frac{8}{y} }} \right.$
$\left \{ {{ \frac{64}{y ^{2} } +y^{2} = 20} \atop {x= \frac{8}{y} }} \right.$
$\left \{y^{4} -20y^{2} + 64= 0 } \atop {x= \frac{8}{y} }} \right.$
Пусть t = y².
$t ^{2} - 20t + 64 = 0$
t₁ + t₂ = 20
t₁*t₂ = 64

t₁ = 4
t₂ = 16
Обратная замена:
y² = 4
y² = 16
$\left \{y = 2 } \atop {x= \frac{8}{y} }} \right.$
$\left \{y = -2 } \atop {x= \frac{8}{y} }} \right.$
$\left \{y = 4 } \atop {x= \frac{8}{y} }} \right.$
$\left \{y = -4 } \atop {x= \frac{8}{y} }} \right.$

$\left \{y = 2 } \atop {x= 4} }} \right.$
$\left \{y = -2 } \atop {x= -4} }} \right.$
$\left \{y = 4 } \atop {x= 2} }} \right.$
$\left \{y = -4 } \atop {x= -2} }} \right.$
Ответ: $(2; 4), (-2; -4), (4; 2), (-4; -2).$

2) $\left \{ {{ x^{2} +y^{2} = 5} \atop {xy=2}} \right.$
$\left \{ {{ x^{2} +y^{2} = 5} \atop {x= \frac{2}{y} }} \right.$
$\left \{ {{ \frac{4}{y ^{2} } +y^{2} = 5} \atop {x= \frac{2}{y} }} \right.$
$\left \{y^{4} -5y^{2} + 4= 0 } \atop {x= \frac{8}{y} }} \right.$
Пусть z = y².
z² - 5z + 4 = 0

z₁ + z₂ = 5
z₁*z₂ = 4

z₁ = 4
z₂ = 1
Обратная замена:
$\left \{y = 1 } \atop {x= \frac{2}{y} }} \right.$
$\left \{y = -1 } \atop {x= \frac{2}{y} }} \right.$
$\left \{y = 2 } \atop {x= \frac{2}{y} }} \right.$
$\left \{y = -2 } \atop {x= \frac{2}{y} }} \right.$

$\left \{y = 1 } \atop {x= 2} }} \right.$
$\left \{y = -1 } \atop {x= -2} }} \right.$
$\left \{y = 2 } \atop {x= 1} }} \right.$
$\left \{y = -2 } \atop {x= -1} }} \right.$
Ответ: $(1; 2), (-1; -2), (2; 1), (-2; -1).$

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