X^3 - 3x^2 - 4x + 12 = 0
x^2*(x-3)-4(x-3)=0
(x^2-4)*(x-3)=0
(x-2)*(x+2)(x-3)=0
x-2=0 x=2
x+2=0 x=-2
x-3=0 x=3
Otvet:
x1=2
x2=-2
x3=3
В) S = ab
P = 2(a + b)
b = S/a = 28/a
P = 2(a + 28a) = 2a + 56/a = (2a^2 + 56)/a
a) P = 2(a + b) = 2a + 2b
b = (P - 2a)/2 = (16 - 2m)/2= 2(8 - m)/2 = 8 - m
S = ab = m(8 - m)
{(2x-y)² -(x+3y)² = -9 ;3x+2y = - 1.
{(2x-y -(x+3y))(2x-y +x+3y) = -9 ; 3x+2y = - 1.
{(2x-y -x-3y)(3x+2y) = -9 ; 3x+2y = - 1.
{(x-4y )(<em>3x+2y</em>) = -9 ; <em>3x+2y</em> = - 1.
{(x-4y )(-1) = -9 ; 3x+2y = - 1.
{x-4y =9 ; 3x+2y = - 1. ||*2||.
{x-4y =9 ; 6x+4y = - 2. * * * сложим * * *
{x-4y =9; 7x=7.
{1 -4y =9 ; x=1⇔{y =-2; x=1.
ответ: (1; -2).
Найти первообразную функции и подтсавить х=2 и х=3
Что-то типо такого
∫_2^3▒〖(-x^2 〗+6x-5)dx=-〖1/3 x〗^3+3x^2-5x=-1/3*3^3+3*3^2-5*3-(-1/3*2^3+3*2^2-5*2)=
<span>и посчитать</span>
x^4 + 12x^2 + 36 - (x^4 - 16x^2 + 64) + 28 =
= x^4 + 12x^2 + 36 - x^4 + 16x^2 - 64 + 28 =
= 28x^2 + 64 - 64 = 28x^2
28x^2 ≡ 28x^2
ч т д