Раскроем модуль
ОДЗ
x > - 1,5
- 3< log2 (2x + 3) < 3
log 2(1/8) < log 2 (2x+ 3) < log2 8
1/8 < 2x +3 < 8
1/8 - 3 < 2x < 8 - 3
- 23/8 < 2x < 5 /:2
- 23/16 < x < 2,5
- 1,4375 < x < 2,5
+ ОДЗ
- 1,4375 < x < 2,5
5/3•5/7–14/3•5/7 5/7(5/3–14/3)
------------------------ = ------------------ =
81/49–9/7•2/7 81/49–18/49
5/7•(-3) 15•49 5
= ---------- = – --------- = – -----
63/49 7•63 3
A1=-1,d=3
a4 = a1+3.d, a4 = -1+3.3=-1+9=8
a8= a1+7d, a8 = -1+7.3=-1+21=20
Formula: an= a1+(n-1).d
3х²-12х+2х-8=0
3х²-10х-8=0
D=(-10)²-4*3*(-8)=100+96=√196=14
x1=10+14/6=24/6=4
x2=10-14/6=-4/6=-2/3
(13/11-13/22)*11/5=
(26/22-13/22)*11/5=
13/22*11/5=13/10=1,3