1000=100%, 1%=10 рублей, 3%=30 рублей, а значит, через два года на 60 рублей.
1
a)-2-(-2)+3=3
b)
![(5 ^{log(5)( \sqrt{2} -1)^2} +3 ^{log(3)( \sqrt{2} +1)^2} )*log(625)5=](https://tex.z-dn.net/?f=%285+%5E%7Blog%285%29%28+%5Csqrt%7B2%7D+-1%29%5E2%7D+%2B3+%5E%7Blog%283%29%28+%5Csqrt%7B2%7D+%2B1%29%5E2%7D+%29%2Alog%28625%295%3D)
![[( \sqrt{2} -1)^2+( \sqrt{2} +1)^2]*1/4=(2-2 \sqrt{2} +1+2+2 \sqrt{2} +1)*1/4=](https://tex.z-dn.net/?f=%5B%28+%5Csqrt%7B2%7D+-1%29%5E2%2B%28+%5Csqrt%7B2%7D+%2B1%29%5E2%5D%2A1%2F4%3D%282-2+%5Csqrt%7B2%7D+%2B1%2B2%2B2+%5Csqrt%7B2%7D+%2B1%29%2A1%2F4%3D)
![4*1/4=1](https://tex.z-dn.net/?f=4%2A1%2F4%3D1)
2
a)(3/2)^x=a
8a²-30a+27=0
D=900-864=36
a1=(30-6)/16=1,5⇒(3/2)^x=3/2⇒x=1
a2=(30+6)/16=9/4⇒(3/2)^x=9/4⇒x=2
b)log(2)x+3log(2)x+3log(2)x=14
7log(2)x=14
log(2)x=2
x=4
3
a)(1/3)^x*(1/9+5/3-1)<7
(1/3)^x*7/9<7
(1/3)^(x+2)<1
x+2>0
x>-2
x∈(-2;∞)
b)log(2)x=a
a²-2a-3<0
a1+a2=2 U a1*a2=-3⇒a1=-1 U a2=3
-1<a<3⇒-1<log(2)x<3⇒1/2<x<8
x∈(1/2;8)
4
1/2*log(3)(6√6-15)²+1/3*log(3)(6√6+15)³=
=log(3)|6√6-15|+log(3)(6√6+15)=log(3)|(6√6-15)(6√6+15)|=
=log(3)|216-225|=log(3)|-9|=2
2=2
A:(b-c)...........................................
Т.к. подкоренное выражение не может быть отрицательным то
(1+2х) \ 5 ≥ 0 ⇒ 2х + 1 ≥ 0 ⇒ 2х ≥ -1 ⇒ х ≥ -1\2
т.е. х∈ [ -1\2 ; + ∞ )
k^2-9+4-4k+k^2-2k^2+4k=-5