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dinonafrie1488
dinonafrie1488
17.12.2021 22:54 •  Алгебра

1.решите уравнение: 6x в квадрате -7x+1=0

👇
Ответ:
kabaramba
kabaramba
17.12.2021
6x^2-7x+1=0
D=49-4*6=5
x1=7+5/12=1
x2=7-5/12=2/12
4,7(41 оценок)
Ответ:
Dirijabl
Dirijabl
17.12.2021
6х^2-7 х+1=0
D=b^2-4ac=(-7)^2-4*6*1=49-24=25
х1=(-b+корень из D)/(2а)=(-(-7)+корень из 25)/(2*6)=(7+5)/12=12/12=1
х2=(-b-корень из D)/(2a)=(-(-7)-корень из 25)/(2*6)=(7-5)/12=2/12=1/6
ответ: 1/6; 1
4,5(4 оценок)
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Ответ:
гсооагвгв
гсооагвгв
17.12.2021

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Must Waste More Time...

4,8(58 оценок)
Ответ:
jarinka2804
jarinka2804
17.12.2021
Составьте уравнение окружности проходящие через точки а (3;13) b(-7;-11) c(10;6)

x-x0)^2+(y-y0)^2=r^2 - общий вид. Подаставляем координаты трех точек:
(1-x0)^2+(2-y0)^2=r^2
x0^2+(1+y0)^2=r^2 (***)
(3+x0)^2+y0^2=r^2

приравняем левые части второго и третьего уравнений:
x0^2+(1+y0)^2=(3+x0)^2+y0^2
xo^2+1+2y0+y0^2=9+6x0+x0^2+y0^2
y0-3x0=4 (*)

теперь приравниваем первое и второе:
(1-х0)^2+(2-y0)^2=x0^2=(1+y0)^2
1-2x0+x0^2+4-4y0+y0^2=x0^2+1+2y0+y0^2
x0=2-3y0 (**)

из уравнений (*) и (**) составляем систему и решаем ее:
у0-6+9у0=4
у0=1
х0= -1

находим радиус, подставив в (***):
(-1)^2+(1+1)^2=r^2; r^2=5. Тогда уравнение окружности:
(х+1)^2+(у-1)^2=5
4,6(7 оценок)
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