Question

Решить неравенство (№4)

Answer 1

2
(x²-2x+6)/(x+1)-x≥0
(x²-2x+6-x²-x)/(x+1)≥0
(6-3x)/(x+2)≥0
x=2  x=-2
x∈(-2;2]
4
1/(x-2)+1/(x-1)-1/x≥0
[x(x-1)+x(x-2)-(x-1)(x-2)]/[x(x-1)(x-2)]≥0
(x²-x+x²-2x-x²+2x+x-2)[x(x-1)(x-2)]≥0
(x²-2)[x(x-1)(x-2)]≥0
x²-2=0⇒x=-√2 U x=√2
x=0
x-1=0⇒x=1
x-2=0⇒x=2
           _                  +                  _                +                _                +
--------------[-√2]----------(0)--------------(1)----------[√2]--------------(2)---------------
x∈[-√2;0) U (1;√2] U (2;∞)

Answer 2

1/(x-2)+ 1/(x-1)≥1/(x)   ⇔1/(x-2)+ 1/(x-1)-1/(x)≥0   ⇔

[x(x-1)+x(x-2)-(x-2)(x-1)]/[x(x-2)(x-1)]≥0 ⇔

[x²-x+x²-2x-x²+3x-2] / [x(x-2)(x-1)]≥0   ⇔   (x²-2)/ [x(x-2)(x-1)]≥0   ⇔
  
          -                           +              -               +            -              +
---------------------(-√2)---////-------(0)-------(1)--////--(√2)-----(2)---////-------------

x∈[-√2;0)∪(1;√2] ∪(2;∞)