Д=25+24= 49
х1=(-5+7)/4=0,5
х2=(-5-7)/4=-3
Cos2x+0.5=cos²x
cos²x-sin²x+0.5=cos²x
sin²x=0.5
sinx=±1/√2
x₁=(-1)^n•π/4+πn
<span>x₂=-(-1)^n•π/4+πn
Какой-то так :)</span>
Это означает, что нужно записать в виде двойного неравенства.![2)x=0,4\pm0,15\\\\0,4-0,15<x<0,4+0,15\\\\0,25<x<0,55\\\\4)x=0,73\pm0,01\\\\0,73-0,01<x<0,73+0,01\\\\0,72<x<0,74\\\\6)x=-2\frac{1}{5}\pm\frac{1}{10}\\\\x=-2,2\pm0,1\\\\-2,2-0,1<x<-2,2+0,1\\\\-2,3<x<-2,1](https://tex.z-dn.net/?f=2%29x%3D0%2C4%5Cpm0%2C15%5C%5C%5C%5C0%2C4-0%2C15%3Cx%3C0%2C4%2B0%2C15%5C%5C%5C%5C0%2C25%3Cx%3C0%2C55%5C%5C%5C%5C4%29x%3D0%2C73%5Cpm0%2C01%5C%5C%5C%5C0%2C73-0%2C01%3Cx%3C0%2C73%2B0%2C01%5C%5C%5C%5C0%2C72%3Cx%3C0%2C74%5C%5C%5C%5C6%29x%3D-2%5Cfrac%7B1%7D%7B5%7D%5Cpm%5Cfrac%7B1%7D%7B10%7D%5C%5C%5C%5Cx%3D-2%2C2%5Cpm0%2C1%5C%5C%5C%5C-2%2C2-0%2C1%3Cx%3C-2%2C2%2B0%2C1%5C%5C%5C%5C-2%2C3%3Cx%3C-2%2C1)
(x-1)²(x-1+5)≥0 (x-1)²(x+4)≥0 <span>x∈<-4,∞)</span>
Х^3-36х=0
Х(х^2-36)=0
Х(х-6)(х+6)=0
Х=0; х=6; х=-6
(Х^2-4)/3 - (5х-2)/6=1 /*6
2х^2-8-5х+10=6
2х^2-5х-12=0
Д=25+8*12=25+96=121
Х1=(5+11)/4=4
Х2=(5-11)/4=1,5
Решение во вложении...