Решите системы уравнений:
[tex] \left \{ {{ \sqrt[3]{x} - \sqrt[3]{y}=2 } \atop {xy=27}} \right.[/tex]
[tex] \left \{ {{ \sqrt{x} + \sqrt{y}=10} \atop { \sqrt[4]{x}+ \sqrt[4]{y} = 4 }} \right. [/tex]
[tex] \left \{ {{ \sqrt[3]{x} - \sqrt[3]{y}=2 } \atop {xy=27}} \right.[/tex]
[tex] \left \{ {{ \sqrt{x} + \sqrt{y}=10} \atop { \sqrt[4]{x}+ \sqrt[4]{y} = 4 }} \right. [/tex]
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ху=27
∛х-∛у=2
х=27/у
∛27/у - ∛у=2
х=27/у
3/∛у -∛у=2 |* ∛y
3- ∛y² =2∛y
∛y² +2∛y -3=0 ∛y=a
a²+2a-3=0
D=4+12=16
a₁=(-2+4)/2=1 ∛y₁=1 y₁=1 x₁=27/1=27 (27; 1)
a₂=(-2-4)/2=-3 ∛y₂=-3 y₂=-27 x₂=27/(-27)= -1 (-1 ;-27)
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√х+√у=10
⁴√х+⁴√у= 4
√х+√у=10 заметим √х= ⁴√х²
⁴√х = 4 -⁴√у
(4 -⁴√у)² +√у=10
16 -8⁴√у+⁴√у² +√у=10
2√у- 8⁴√у +16=10
2√у- 8⁴√у +6=0 ⁴√у =а ОДЗ а>0
2а² -8а+6= 0
D=64-48=16
a=(8+4)/4=3 ⁴√у₁ =3 y₁=3⁴=81 ⁴√х₁= 4 -⁴√у = 4-3=1 x₁= 1 (1; 81)
a₂=(8-4)/4=1 ⁴√у₂ =1 y₂=1 ⁴√х₂= 4 -⁴√у =4- 1=3 x₂=3⁴=81 (81; 1)