Question

Помогите с системами уравнений пожалуйста

Answer 1

tex]b) left { {{2 sqrt{x} - 3 sqrt{y} = 1} atop {3 sqrt{x} - 2 sqrt{y} = 4}} right. [/tex]
Пусть $a = sqrt{x}, b = sqrt{y}.$
$left { {{2a - 3b = 1} atop {3a - 2b = 4}} right. \ \ left { {{4a - 6b = 2} atop {3a - 2b = 4}} right. \ \ 5a = -10 \ left { {{a = 5} atop {2*2 - 3b = 1}} right. \ \ left { {{a = 2} atop {b = 1}} right.$
Обратная замена:
$left { {{ sqrt{x} = 2} atop { sqrt{y} = 1}} right. \ \ left { {{x = 4} atop {y = 1}} right. \ \ boxed{OTBET: (4; 1).}$

$r) left { {{ sqrt[3]{x} + sqrt[4]{y} = 3} atop {3 sqrt[3]{x}-5 sqrt[4]{y} =1 }} right.$
Пусть $a = sqrt[3]{x}, b = sqrt[4]{y} .$
$left { {{a + b = 3} atop {3a - 5b = 1}} right. \ \ left { {{b = 3 - a} atop {3a - 5(3 - a) = 1}} right. \ \ left { {{b = 3 - a} atop {3a = 15 + 5a = 1}} right. \ \ left { {{8a = 16} atop {b = 3 - a}} right. \ \ left { {{a = 2} atop {b = 1}} right.$
Обратная замена:
$left { {{ sqrt[3]{x} = 2 } atop { sqrt[4]{y} = 1 }} right. \ \ left { {{x = 2} atop {y=1}} right. \ \ boxed{ OTBET: (8; 1)}$

Answer 2

б)
{ 2√x - 3√y = 1 ; 3√x - 2√y = 4 ⇔ || 1-ое  урав.*(-2) , 2-ое _*(3)  ||  
{ - 4√x+ 6√y = -2 ; 9√x - 6√y = 12 ⇔ || + ||
{ 5√x = 10 ;  2√x - 3√y = 1 ⇔ { √x =2 ;  2*2  - 3√y = 1 .⇔ {√x =2 ;  √y =1. 
⇒ { x =4 ; y =1.
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г)
 { ∛x  + y^(1/4) =3 ;  3∛x  - 5* y^(1/4) =1. ⇔|| 1-ое  урав.*(5)  и "+" || 
{  8∛x  =16   ; ∛x  + y^(1/4) =3. ⇔{ ∛ =2  ;  2+ y^(1/4)=3. ⇔ { ∛ =2 ; y^(1/4)=1 ⇒ { x =8 ;  y =1.