А)f`(x)=1/2*1/x²
б)f`(x)=-2/x² -2
в)f`(x)=[(2x-2)(x+1)-(x²-2x)]/(x+1)²=(2x²+2x-2x-2-x²+2x)/(x+1)²=
=(x²+2x-2)/(x+1)²
г)f`(x)=[(3x²-2)(x²+2)-2x(x³-2x+3)]/(x²+2)²=
=(3x^4+6x²-2x²-4-2x^4+4x²-6x)/(x²+2)²=(x^4+8x²-6x-4)/(x²+2)²
Х/х-у :(х+у)(х-у)+у*у/(х-у)у = х/х-у : x^2-y^2+y^2
/(х-у)у = х/х-у : x^2
/(х-у)у = x*y(x-y)/(x-y)*x^2 = y/x
Ответ:[-7/24π+πn/2; π/24+πn/2],n∈z
Объяснение: 2sin2x·cos2x ≤1/4 ·2;
sin4x ≤1/2;
-7/6 π+2πn≤4x≤π/6 +2πn, n∈z;
-7/24 π+πn/2≤x≤π/24+πn/2,n∈z;
3x=одна девяносто девятая
X= 33