Question

решите системы,пжпжпжпж ДАЮ 30 БАЛЛЛОВ

5)4х^2+y=9

8x^2-y=3

6) 3x^2+y=4

2x^2-y=1

7)2x^2+y=4

4x^2-y=2

8)5x^2+y=12

9x^2-y=2

Answer 1

$5)+\left \{ {{4x^{2}+y=9 } \atop {8x^{2}-y=3 }} \right.\\ --------\\12x^{2}=12\\\\x^{2}=1\\\\x_{1} =-1\\\\x_{2}=1$

$\left \{ {{x=-1} \atop {y=9-4*(-1)^{2} }} \right. \\\\\left \{ {{x=-1} \atop {y=5}} \right.\\\\\left \{ {{x=1} \atop {y=9-4*1^{2} }} \right.\\\\\left \{ {{x=1} \atop {y=5}} \right.$

Ответ : (- 1 ; 5) , (1 ; 5)

$6)+\left \{ {{3x^{2}+y=4 } \atop {2x^{2}-y=1 }} \right.\\\\-------\\5x^{2}=5\\\\x^{2}=1\\\\x_{1}=-1\\\\x_{2}=1\\\\y_{1}=4-3*(-1)^{2}=1\\\\y_{2}=4-3*1^{2} =1$

Ответ : (- 1 ; 1) , (1 ; 1)

$7)+\left \{ {{2x^{2}+ y=4} \atop {4x^{2}-y=2 }} \right.\\ --------\\6x^{2} =6\\\\x^{2}=1\\\\x_{1}=-1\\\\x_{2}=1\\\\y_{1} =4-2*(-1)^{2}=2\\\\y_{2}=4-2*1^{2}=2$

$8)+\left \{ {{5x^{2}+y=12 } \atop {9x^{2}-y=2 }} \right.\\ --------\\14x^{2}=14\\\\x^{2}=1\\\\x_{1}=-1\\\\x_{2}=1\\\\y_{1}=12-5*(-1)^{2}=7\\\\y_{2}=12-5*1^{2}=7$

Ответ : (- 1 ; 7) , ( 1 ; 7)