А/(a-b) - b/a - 1 = (a*a-b(a-b)-1(a(a-b))/(a(a-b)) = (a²-ab+b²-a²+ab)/(a(a-b)) = b²/(a²-ab)
3tg2x=-√3
tg2x=-√3/3
2x=-π/6+πn
x=-π/12+πn/2,n∈z
4*a^3*b* 12*(ab^2)^3 * 3a =
= 4*a^3*b * 12a^3*b^6 * 3a =
= 4*12*3*a^3*a^3*a*b*b^6 =
= 144a^(3 + 3 + 1) * b^(1 + 6) =
= 144*a^7*b^7
Дано: F(x)=0.3x^10+2x^7-4x<span>, f(x)=3x^9+14x^6-4
F`(x)=(0.3x^10)`+(2x^7)`-(4x)`
F`(x)=3x^9+14x^6-4
</span>3x^9+14x^6-4=3x^9+14x^6-4
<span>F`(x)=f(x), значит:
</span>F(x)=0.3x^10+2x^7-4x - первообразная для f(x)=3x^9+14x^6-4, что и требовалось доказать.<span>
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