Помогите с системами уравнений пожалуйста

Question 1

Question 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.

Question 3

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:
$\left \{ {{ \sqrt[3]{x} = 2 } \atop { \sqrt[4]{y} = 1 }} \right. \\ \\ \left \{ {{x = 2} \atop {y=1}} \right. \\ \\ \boxed{ OTBET: (8; 1)}$

oganesbagoyan_zn · Answer 1 · 2018-05-11T03:29:32+0000

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.
----------
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.