| x+5 | ||
>0 | ||
| 4−x |
| 3x−15 | |
<0 | |
| x−2 |
| 2x+1 | |
>1 | |
| x+2 |
| 5x−1 | |
<1 | |
| x+6 |
| x+5 | ||
a) | >0 i 4−x≠ 0, otóż iloraz ten będzie dodatni ⇔ (x+5)(4−x) >0 i −x≠ −4 ⇔ | |
| 4−x |
| 3x−15 | ||
b) | < 0 i x−2≠ 0 ⇔ 3(x−5)(x−2)< 0 i x≠ 2 ⇔ x∊(2; 5) , | |
| x−2 |
| 2x+1 | ||
c) | < 1 /*(x+2)2>0 ⇔ (2x+1) (x+2)< (x+2)2 i x+2≠0 ⇔ | |
| x+2 |
−−−−−−−−−
d) analogicznie jak c)
| 5x−1 | |
<1 i x nie rowna sie (−6) | |
| x+6 |
| 5x−1 | |
−1<0 | |
| x+6 |
| 5x−1−(x+6) | |
<0 | |
| x+6 |
| 4x−7 | |
<0 | |
| x+6 |
| 7 | 7 | |||
4(x− | )(x+6)<0 to x nalezy (−6, | ) | ||
| 4 | 4 |