-2х+3(1+х)≤3х+1
-2х+3+3х-3х-1≤0
-2х+2≤0
-2х≤-2
х≥1
Ответ: х∈[1; + бесконечность).
=3m×(n^2-16)
=3m×(n-4)+(n+4)
<span>3m^2(m+2n)-2n(8m^2-n)=3m^3+6m^2n-16m^2n+2n^2=3m^3-10m^2n+2n^2</span>
3x + m = 0
3x = - m
x= - m/3
Cosα+2cos3α+cos5α=(cosα+cos5α)+2cos3α=
=2cos((α+5α)/2)*cos((5α-α)/2)+2cos3α=
=2cos3α*cos2α+2cos3α=2cos3α(cos2α+1)=
=2cos3α(cos²α-sin²α+cos²α+sin²α)=
=2cos3α*2cos²α=4cos3α*cos²α