A) 6 - 16)(x2 + 1) = 0; 6) (x2 + 9)(x2 – 4) 0); B) (25 – x2)(x + 6)2 0; - id1819409520220416 от Roman310399 16.04.2022 14:27

Question

Алгебра: A) 6 - 16)(x2 + 1) = 0; 6) (x2 + 9)(x2 – 4) 0); B) (25 – x2)(x + 6)2 0;​

Roman310399 · Accepted Answer

1) y = 8√x + 3x^5 ?    2) y = 5x² +3(1/x -4) ; 3) y = (x^4)/ (3-x) или y =x^4/3 -   x    ?
1) y '= (8√x + 3x^5 )' = (8√x ) '+ (3x^5)' =8(√x) + 3(x^5)' =8*1/2*(x^(-1/2)) +3*5*x^4=
=4/√x +15x^4.
2) у=(5х² +3(1/x-4))' =(5х² +3/x- 12) ' = (5х²) ' +(3/x) - (12) ' =5*(х²) ' +3*(1/x) - 0 =
5*2x +3(-1/x²) =10x -3/x² .
3)
3a) y '= ((x^4)/(3-x) ) =((x^4)' * (3-x) -(x^4)*(3 -x)')/(3-x)² =((4x³(3 -x) - (x^4)*(-1))/(3-x)²
=(12x³ - 4x^4 + x^4)/(3-x)² =(12x³ -3x^4)/(x-3)² =3x³(4-x)/(x-3)² .

3b)   y ' =(x^4/3 - x ) = (x^4/3) - (x ) ' =4/3*(x^1/3) -1 =4/3*∛x -1.

A) 6 - 16)(x2 + 1) = 0;
6) (x2 + 9)(x2 – 4) > 0);
B) (25 – x2)(x + 6)2 > 0;

MOGZ ответил

A) 6 - 16)(x2 + 1) = 0;6) (x2 + 9)(x2 – 4) > 0);B) (25 – x2)(x + 6)2 > 0;​

MOGZ ответил

A) 6 - 16)(x2 + 1) = 0;
6) (x2 + 9)(x2 – 4) > 0);
B) (25 – x2)(x + 6)2 > 0;