Răspuns:
c. x≥0
d. x > [tex]\frac{-3}{5}[/tex]
e. x∈[-4,-3/2) U[-1,∞)
Explicație pas cu pas:
c. [tex]\frac{2}{x-1} + \frac{5x-3}{x-1}[/tex]≥1 / *(x-1)
2+5x-3≥x-1
5x-x≥-1+3-2
4x≥0
x≥0
d. [tex]\frac{1}{2(x+2)} -\frac{2x+1}{x+2}[/tex] < [tex]\frac{1}{2}[/tex] /*2(x+2)
1-2(2x+1) < x+2
1-4x-2 < x+2
-4x-x < 2+2-1
-5x < 3 /:(-5)
x > [tex]\frac{-3}{5}[/tex]
e. [tex]\frac{(x+4)(x+1)}{2x+3}[/tex] > 0
x | -∞ -4 -3/2 -1 +∞
x+4 |------------0+++++++++++++++++++++
x+1 | ------------------------------------0++++++
2x+3|------------------------0++++++++++++++
ect. |------------0+++++++/------------0+++++
x+4=0
x= -4
x+1=0
x= -1
2x+3=0
x= -3/2
x∈[-4,-3/2) U[-1,∞)
