1) х²-4х+3 ≥ 0
х²-4х+3=0
D= 16-12 = 4
х₁ = 4-2 / 2 = 1
х₂ = 4+2 / 2 = 3
+ - +
--------.---------.-------->х
1 3
х∈(-∞; 1]∪[3; +∞)
2) 5х²-11х > -2
5x²-11x+2 > 0
D = 121-40 = 81
х₁ = 11-9 / 10 = 2/10 = 1/5
х₂ = 11+9 / 10 = 2
+ - +
-----------₀----------₀--------->x
1/5 2
x∈(-∞; 1/5)∪(2; +∞)