Question

Помогите пожалуйста)

Answer 1

∛(m³-6m²n+12mn²-8n³)/(2n-m)=∛(m³-3*m²*2n+3*m*(2n)²-(2n)²)/(2n-m)=
=∛(m-2n)³/(2n-m)=(m-2n)/(2n-m)=-(m-2n)/(m-2n)=-1.

log₄64-log₄(1/16)-3*log₄2=log₄2⁶-log₄2⁻⁴-log₄2³=log₄(2⁶/(2⁻⁴*2³)=
=log₄(2⁶/2⁻¹)=log₄2⁷=7*log₄2=7*log₂²2=(7/2)*log₂2=3,5.

(1/4)ˣ+(1/4)ˣ⁻¹>5
(1/4)ˣ+(1/4)ˣ/(1/4)>5
(1/4)ˣ+4/(1/x)>5
5*(1/4)ˣ>5  |÷5
(1/4)ˣ>1
(1/4)ˣ>(1/4)⁰
Так как (1/4)<0   ⇒
x<0.

Answer 2

tg3π/4 + cos5π/3 ·sin7π/6 - (sin2π/3)²=
= -1+ 1/2 · (-1/2) - (√3/2)²= -1 - 1/4 - 3/4 = - 2 

(¹/₄)ˣ + (¹/₄)ˣ⁻¹ > 5
4⁻ˣ + 4⁻ˣ⁺¹ > 5
4⁻ˣ·(4⁰+4¹)>5
4⁻ˣ·5>5
4⁻ˣ>1
4⁻ˣ>4⁰   основание 4 больше 1, ⇒
-х>0
x<0
x∈(-∞;0)