срочно нужна!!!(ответы с полными решениями!!!). Created by Jarvis2002. algebra-ru

(*) (7+y)y = -127y + y² + 12 = 0y1 = -3 y2 = -4 1) 2) (*) (4 + y)² - y² = 4016 + 8y + y² - y² = 408y = 24y = 3(*) (4 + y)y = 124y + y² = 12y² + 4y -12 = 0y1 = 2y2 = -61) 2) (*) (5 - y)y = -14y² - 5y -14 = 0y1 = -2y2 = 71) 2)

срочно нужна!!!(ответы с полными решениями!!!)...

Jarvis2002 @Jarvis2002

July 2022 1 9 Report

срочно нужна!!!(ответы с полными решениями!!!)

Answers & Comments

Etzel $\left \{ {{x-y=7} \atop {xy=-12}} \right. \\ \left \{ {{x=7+y} \atop {(7+y)y=-12(*)}} \right.$

(*) (7+y)y = -12
7y + y² + 12 = 0
y1 = -3
y2 = -4

1) $\left \{ {{x=7+y} \atop {-3}} \right. \\ \left \{ {{x=7-3} \atop {-3}} \right. \\ \left \{ {{x=4} \atop {y=-3}} \right.$
2) $\left \{ {{x=7+y} \atop {y=-4}} \right. \\ \left \{ {{x=7-4} \atop {y=-4}} \right. \\ \left \{ {{x=3} \atop {y=-4}} \right.$

$\left \{ {{x-y=4} \atop {x^2-y^2=40}} \right. \\ \left \{ {{x=4+y} \atop {(4+y)^2-y^2=40(*)}} \right.$
(*) (4 + y)² - y² = 40
16 + 8y + y² - y² = 40
8y = 24
y = 3

$\left \{ {{x=4+y} \atop {y=3}} \right. \\ \left \{ {{x=7} \atop {y=3}} \right.$

$\left \{ {{x-y=4} \atop {xy=12}} \right. \\ \left \{ {{x=4+y} \atop {(4+y)y=12(*)}} \right.$
(*) (4 + y)y = 12
4y + y² = 12
y² + 4y -12 = 0
y1 = 2
y2 = -6

1) $\left \{ {{x=4+y} \atop {y=2}} \right. \\ \left \{ {{x=6} \atop {y=2}} \right.$
2) $\left \{ {{x=4+y} \atop {y=-6}} \right. \\ \left \{ {{x=-2} \atop {y=-6}} \right.$

$\left \{ {{x+y=5} \atop {xy=-14}} \right. \\ \left \{ {{x=5-y} \atop {(5-y)y=-14(*)}} \right.$
(*) (5 - y)y = -14
y² - 5y -14 = 0
y1 = -2
y2 = 7
1) $\left \{ {{x=5-y} \atop {y=-2}} \right. \\ \left \{ {{x=7} \atop {y=-2}} \right.$
2) $\left \{ {{x=5-y} \atop {y=7}} \right. \\ \left \{ {{x=-2} \atop {y=7}} \right.$

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