(x-1)²<√2(x-1)
x-1=t ⇒
t²-(√2)t<0
t(t-√2)<0
-∞______+______0______-_______√2_______+______+∞
0<t<√2
0<x-1<√2
1<x<1+√2 ⇒
x∈(1;1+√2)
3x^2 - 5x - 2 = 0
x1 = 2
x2 = - 2/6
x1 + x2 = 2 + ( - 2/6) = 2 - 2/6 = 10/6 = 5/3 = 1целая 2/3
x1 * x2 = - 2/6 * 2 = - 4/6 = - 2/3
3х-2(х-5)≥-6
3х-2х+10<span>≥-6
3х-2х</span><span>≥-6-10
х</span><span>≥-16</span>
я немогу помочь но вот сайт- wolframalpha.com