(6х-12)×(0,5-4)= 0
3х - 24х - 6 + 48 = 0
-21х+42=0
-21х=-42
х= 2
ответ: х=2
Объяснение:
а) х² - 8х = 0, х·(х -8) = 0 ⇒ х =0 или х - 8 = 0; х =0 или х = 8.
б. 6х² = 12; х² = 12÷6, х² = 2, х = ±√2
в) 3x² – 48 = 0, 3x²= 48, x² = 48÷3,x² = 16, х = ± 4
г) 6x² – 5x + 1 = 0;D = b²- 4ac = 25 - 4·6 = 24; x = -b ±√D/2a
x1 = 5+√1/12 = 5+1/12 = 6/12 = 1/2, x2 = 5-1/12 = 4/12 = 1/3
д) x² –16x + 71 = 0.D = b²- 4ac =256 - 4·1·71= 256 -284 =-28 - меньше 0 ⇒∅
е) (4x – 3)2 + (3х – 1)(3х+1) = 9
8х -6 +(9х²-3х+3х-1)=9; 8х -6+(9х²-1) =9; 8х -6 +9х²-1-9 = 0; 9х²+8х-16 =0
D = b²- 4ac = 64+4·9·16= 64+576 =640
х1 = -8+√640/18/= -8+8√10/18; х2 = -8-8√10/18
2*.При яких значеннях а рівняння аx² + аХ + 36 = 0 має один корінь?
D = 0⇒ а²-4·а·36 = 0, а²-144 = 0, а²=144, а = ±12
x + y = П/4
sinx/cosx + siny/cosy = 1 | x,y <> П/2 + Пk
sinx*cosy + siny*cosx = cosx*cosy
sin(x+y) = cosx*cosy
cosx*cosy = sin(П/4)
cosx*cos(П/4-x) = sin(П/4)
cosx*(cos(П/4)*cos(x) + sin(П/4)*sin(x)) = sin(П/4) | cos(П/4) = sin(П/4)
cosx*(cosx+sinx) = 1
cos^2x + cosx*sinx = 1
cosx*sinx - sin^2x = 0
sinx*(cosx - sinx) = 0
sinx = 0 -> x = Пk, y = П/4 - Пk
cosx = sinx -> x = П/4 - Пk, y = Пk
cos^2x = sinx*siny
sin^2x = cosx*cosy
1 = sinx*siny + cosx*cosy
1 = cos(x-y)
x-y = П/2 + 2Пk, y = x + П/2 + 2Пk
cos^2x = sinx*sin(x+П/2) = sinx*cosx -> cosx = 0 | cosx = sinx
sin^2x = cosx*cos(x+П/2) = cosx*(-sinx) -> sinx = 0 | sinx = -cosx
--> cosx = 0 | sinx = 0 --> x = Пn/2, y = П(n+1)/2 + 2Пk
cosx*sqrt(cos2x) = 0 | cos2x >= 0
2sin^2x = cos(2y-П/3) | 2sin^2x <= 1
cosx*sqrt(cos^2x - sin^2x) = 0
cosx*sqrt(1 - 2sin^2x) = 0
cosx*sqrt(1 - cos(2y-П/3)) = 0
cosx = 0 -> x = П/2 + Пk - > 2sin^2x > 1 - не подходит
cos(2y-П/3) = 1 - > 2y - П/3 = П/2 + 2Пk -> y = 5П/12 + Пk | cos2x = 1 - 2sin^2x = 1 - cos(2y-П/3) = 0 -> x = П/4 + Пn/2
--> x = П/4 + Пn/2, y = 5П/12 + Пk/2
Объяснение:
x=2
Объяснение:
6x-12=0
6x=12
x=2