Х(х-3)=0
Откуда х=0 и х-3=0, х=3
Ответ: х=0 и х=3
2cos2acosa+cos2a=cos2a(2cosa+1)=cos7π/3(2cos7π/6+1)=
=cos(2π+π/3)(2cos(π+π/6)+1)=cosπ/3(-2cosπ/6+1)=1/2(-√3+1)=(1-√3)/2
2.7*(1.7³-1.5³)/(1.5²+5.1*4.5+4.5²)=2.7(1.7-1.5)(1.7²+1.7*1.5+1.5²)/(1.5²+5.1*4.5+4.5²)=2.7*0.2*(2.89+2.55+2.25)/(2.25+22.95+20.25)=
=4.1526/27.45=0.1513