Question

помогите,пожалуйста.Буду благодарна

Answer 1

(ctgα+1)²+(ctgα-1)²=(cosα/sinα+1)²+(cosα/sinα-1)²=
=((cosα+sinα)/sinα)²+((cosα-sinα)/sinα)²=(cosα+sinα)²/sin²α+(cosα-sinα)/sin²α=
=(cos²α+2*sinα*cosα+sin²α)/sin²α+(cos²α-2*sinα*cosα+sin²α)/sin²α=
=(1+2*sin²α*cos²α)/sin²α+(1-2*sinα*cosα+sin²α)/sin²α=
=(1+2*sinα*cosα+1-2*sinα*cosα)/sin²α≡2/sin²α.

(5/3)⁵ˣ⁺²<(3/5)³ˣ⁻¹⁰
(5/3)⁵ˣ⁺²<(5/3)¹⁰⁻³ˣ
5x+2<10-3x
8x<8  |÷8
x<1.
Ответ:  x∈(-∞;1).

4ˣ+2ˣ⁺¹-6=0
2²ˣ+2*2ˣ-6=0
Пусть 2ˣ=t>0  ⇒
t²+2t-6=0   D=28
t₁=-1-√7 ∉
t₂=-1+√7
2ˣ=-1+√7
log₂2ˣ=log₂(-1+√7)
x=log₂(-1+√7).