ОДЗ 5-x≥0⇒x≤5
4cos2x=1
cos2x=1/4
2x=+-arccos1/4+2πn,n∈z
x=+-1/2arccos1/4+πn,n∈z
x1=-1/2arccos1/4
x2=1/2arccos1/4
x3=-π/2-1/2arccos1/4
x4=π/2+1/2arccos1/4
x5=5
1)-8х=3.2
х=3.2÷(-8)
х=0.4
2)2/3х=6
х=6÷2/3
х=6/1×3/2
х=9/2
3)4-5х=0
-5х=4+0
х=-0.8
4)10х+7=3
10х=3-7
10х=-4
х=-0.4
#2
a)75a^4b^3c
#3
15^2a^4b^2)
#4
5^3a^3b^6
#5
а)-27а^3b^3 25b^2=-675a^3b^5
(Чем смогла,тем помогла))