Все решение ниже на фото:
F`(x)=-2/x³-cosx
F`(x)=-2/x³-cosx≠<span> f(x) =-(1/x^3) -cos(x)⇒не является
F`(x)=2x-1
</span><span>F`(x)=2x-1=f(x)=2x-1⇒является</span>
25х - х² = 0
-x² + 25x = 0
а= - 1 ; b = 25 ; с= 0
D = 25² - 4*(-1)*0 = 25² = 625
x₁ = ( - 25 - 25)/(2*(-1)) = - 50/(-2) = 25
x₂ = (-25+25)/( 2 *(-1)) = 0
3x² - 48 = 0
a=3 ; b=0 ; с= - 48
D = 0 - 4*3*(-48) = 576 = 24²
x₁ = -24/(2*3) = - 24/6 = - 4
x₂ = 24/(2*3) = 24/6 = 4
27х - х² = 0
-х² + 27х = 0
а= - 1 ; b = 27 ; с =0
D = 27² - 4*(-1)*0 = 27² = 729
x₁ = ( -27 - 27)/(2*(-1)) = -54/ -2 = 27
x₂ = (-27 + 27) / ( 2*(-1)) = 0
4x² - 36 = 0
a= 4 ; b = 0 ; с = -36
D = - 4*4*(-36) = 576 = 24²
x₁ = -24/(2*4) =- 24/8 = - 3
х₂ = 24/(2*4) = 24/8 = 3
1)а^2-12а+36-36-5а=а^2-17а=а(а-17)
2)2х^2-2ху-3х-36