Производная y'(x)=-2+x²-3*x³ в точке x0=1 принимает значение y'(x0)=-2+1²-3*1³=-4. Ответ: y'(1)=-4.
(a-x)(x³-y³)-(x-y)(a³-x³)=
=(a-x)(x-y)(x²+xy+y²)-(x-y)(a-x)(a²+ax+x²)=
=(a-x)(x-y)(x²+xy+y²-a²-ax-x²)=
=(a-x)(x-y)(y²-a²+xy-ax)
2x³-2xy²-6x²+6y²=
=(2x³-2xy²)-(6x²-6y²)=
=2x(x²-y²)-6(x²-y²)=
=(x²-y²)(2x-6)=
=2(x-3)(x-y)(x+y)
5a²-5b²-10a³b+10ab³=
=(5a²-5b²)-(10a³b-10ab³)=
=5(a²-b²)-10ab(a²-b²)=
=(a²-b²)(5-10ab)=
=5(1-5ab)(a-b)(a+b)
36x³-144x-36x²+144=
=(36x³-36x²)-(144x-144)=
=36x²(x-1)-144(x-1)=
=(x-1)(36x²-144)=
=(x-1)(6x-12)(6x+12)=
=(x-1)*6(x-2)*6(x+2)=
=36(x-1)(x-2)(x+2)
y³+ay²-b²y-b²a=
=(y³+ay²)-(b²y+b²a)=
=y²(y+a)-b²(y+a)=
=(y+a)(y²-b²)=
=(y+a)(y-b)(y+b)
= ( 3X^2 + X^2 ) + ( - 5X + 2X + 3X) + ( - 4 - 1 ) =
= 4X^2 - 5
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X = 1/2 = 0.5
4 * ( 0.5)^2 - 5 = 4 * 0.25 - 5 = 1 - 5 = ( - 4 )
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ответ ( - 4 )
<span>2х² · 7ху² - 4х²у (-ху) - 3х · 5х²у² = 14 х</span>³<span>у</span>²<span> - 4х</span>³<span>у</span>²<span> - 15х</span>³<span>у</span>²<span> = -5х</span>³<span>у</span>²
-5х³у² х = -3, у=5
-5 . (-3)³ . 5² = 3375
(-3)³ = -27
5² = 25
-5 . (-27) = 135
135 . 25 = 3375