Question

РЕБЯТА,ПОМОГИТЕ,СРОЧНО!!! 100 БАЛЛОВ! 1,2,3,4,5 номера!!!

Answer 1

1.(1;2)
2.$\left \{ {{5x+y=61} \atop {5x-y=-11}} \right.$
10x=50
x=5
$\left \{ {{x=5} \atop {y=5*5+11}} \right.$$\left \{ {{y=36} \atop {x=7,2}} \right.$
Ответ: (5;36)3.
$\left \{ {{ x^{2} + y^{2} =61} \atop { x^{2} - y^{2} =-11}} \right.$
Пусть y²-y', x²-x' тогда
$\left \{ {{x'+y'=61} \atop {x'-y'=-11}} \right.$
2x'=50
$\left \{ {{x'=25} \atop {y'=61-25}} \right.$$\left \{ {{y'=36} \atop {x'=25}} \right.$
x=√x'
$\left \{ {{x= \sqrt{x'} } \atop {y= \sqrt{y'}}} \right.$$\left \{ {{y=6} \atop {x=5}} \right.$ или $\left \{ {{y=-6} \atop {x=-5}} \right.$
Ответ(5;6),(-5;-6).
4. Ответ: 12
5.$\left \{ {{y= x^{2} -36} \atop { x^{2} + y^{2} =36 }} \right.$
$\left \{ {{y-x^{2} =-36} \atop { x^{2} + y^{2} =36 }} \right.$
y²+y=0
y(y+1)=0
y"=-1

$\left \{ {{y'₁=0} \atop {x'₁=6}} \right. \left \{ {{y'₂=0} \atop {x'₂=-6}} \right.$

$\left \{ {{y"₁=-1} \atop {x"₁= \sqrt{36-1} }} \right.$
$\left \{ {{y"₂=-1} \atop {x"= \sqrt{35 }} \right. \left \{ {{y"₂=-1} \atop {x"=- \sqrt{35} }} \right.$
Ответ:(0;6), (0;-6) , (-1, √35) ,(-1; -√35)