1) y'=(3x+x^3)' (x-2) + (3x+x^3) (x-2)' = (3+3x^2)(2-x)+ 3x + x^3 = 6 -3x + 6x^2 - 3x^3 + 3x + x^3 = - 2x^3 + 6x^2 + 6
<span>у=log₂₀(x²-x)
</span>log₂₀(x²-x)>0
x²-x>1
x²-x-1 = (x-1/2+√5/2)(x-√5/2-1/2)>0
x∈(-∞, 1/2-√5/2)⋃(√5/2+1/2, ∞)
Х-у=4
ху+у²=6
х=4+у
(4+у)у+у²=6
2у²+4у-6=0 |:2
у²+2у-3=0
D=4-4*1*(-3)=4+12=16
√D=4
Y1=(-2+4)/2=1
Y2=(-2-4)/2=-3
X1=5
X2=1
<u>2х-1 </u> - <u>х+1</u> = 1
5 2
<u>4х-2-5х-5</u> = 1<u>
</u> 10
-х=10+7
х=-17