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prostatak
@prostatak
December 2021
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Равносильные преобразования неравенств
[tex] \sqrt[3]{x^3-3x^2+2x+8}< 1+x[/tex]
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volkodav575
Verified answer
Возведем в куб
х³-3х²+2х+8<1+3x+3x²+x³
х³-3х²+2х+8-1-3x-3x²-x³<0
-6x²-x+7<0
6x²+x-7>0
D=1+168=169 √D=13
x1=(-1-13)/12=-14/12=-7/6
x2=(-1+13)/12=12/12=1
+ _ +
-----------------------------------------------------
-7/6 1
x∈(-≈;-7/6)U(1;≈)
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Verified answer
Возведем в кубх³-3х²+2х+8<1+3x+3x²+x³
х³-3х²+2х+8-1-3x-3x²-x³<0
-6x²-x+7<0
6x²+x-7>0
D=1+168=169 √D=13
x1=(-1-13)/12=-14/12=-7/6
x2=(-1+13)/12=12/12=1
+ _ +
-----------------------------------------------------
-7/6 1
x∈(-≈;-7/6)U(1;≈)