√3+2cosx=0
2cosx=-√3
cosx=-√3/2
x=+-(π-arccos√3/2)+2πn, n∈Z
x=+-(π-π/6)+2πn, n∈Z
x=+-5π/6+2πn, n∈Z
(x-a)(c-b)+2ab=4x+ac
xc-ac+ab-xb+2ab=4x+ac
x(c-b-4)=2ac-3ab
x(c-b-4)=a(2c-3b)
x=a(2c-3b)/(c-b-4)
2√3(2-5√12)=2·2√3-2·5√12·√3=4√3-10√36=4√3-10·6=4√3-60
(9/5)²+28,8/7*8*77*8/16+1,8
81/25+144/5 / 7*8*77*8/17,8
801/25/7*8*77*8/89/5
801/862400/89/5
9/172480
3+400+12000
12403