5.
jūnijā
Eksāmens MATEMĀTIKĀ 9. KLASEI
Trenējies ŠEIT!

### Teorija

Nevienādības, kurās reizinājums vai dalījums salīdzināts ar nulli ir, piemēram, $\left(x+3\right)\left(x-2\right)>0;\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\frac{x+3}{x-5}\le 0$.

Viena no šādu nevienādību atrisināšanas metodēm ir aizstāšana ar nevienādību sistēmām.

Lai  nevienādību aizstātu ar nevienādību sistēmu, jāzina zīmju likumi:

$\begin{array}{l}\left(+\right)\cdot \left(+\right)=\left(+\right)\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left(-\right)\cdot \left(-\right)=\left(+\right)\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left(+\right)\cdot \left(-\right)=\left(-\right)\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left(-\right)\cdot \left(+\right)=\left(-\right)\\ \\ \frac{\left(+\right)}{\left(+\right)}=\left(+\right)\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.294em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\frac{\left(-\right)}{\left(-\right)}=\left(+\right)\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.441em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\frac{\left(+\right)}{\left(-\right)}=\left(-\right)\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.294em}{0ex}}\phantom{\rule{0.147em}{0ex}}\frac{\left(-\right)}{\left(+\right)}=\left(-\right)\end{array}$

Lai reizinājums būtu pozitīvs, abiem reizinātājiem jābūt ar vienādām zīmēm - vai nu pozitīviem vai negatīviem.
Lai reizinājums būtu negatīvs, reizinātājiem jābūt ar dažādām zīmēm.

$\begin{array}{l}f\left(x\right)\cdot g\left(x\right)>0\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}⇔\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left\{\begin{array}{l}f\left(x\right)>0\\ g\left(x\right)>0\end{array}\right\\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\mathit{vai}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left\{\begin{array}{l}f\left(x\right)<0\\ g\left(x\right)<0\end{array}\right\\\ \\ f\left(x\right)\cdot g\left(x\right)<0\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}⇔\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left\{\begin{array}{l}f\left(x\right)>0\\ g\left(x\right)<0\end{array}\right\\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\mathit{vai}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left\{\begin{array}{l}f\left(x\right)<0\\ g\left(x\right)>0\end{array}\right\\\ \\ f\left(x\right)\cdot g\left(x\right)\ge 0\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}⇔\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left\{\begin{array}{l}f\left(x\right)\ge 0\\ g\left(x\right)\ge 0\end{array}\right\\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\mathit{vai}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left\{\begin{array}{l}f\left(x\right)\le 0\\ g\left(x\right)\le 0\end{array}\right\\\ \\ f\left(x\right)\cdot g\left(x\right)\le 0\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}⇔\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left\{\begin{array}{l}f\left(x\right)\ge 0\\ g\left(x\right)\le 0\end{array}\right\\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\mathit{vai}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left\{\begin{array}{l}f\left(x\right)\le 0\\ g\left(x\right)\ge 0\end{array}\right\\end{array}$

Dalījuma likumi ir gandrīz tādi paši, kā reizinājuma likumi.

Lai dalījums būtu pozitīvs, skaitītājam un saucējam jābūt ar vienādām zīmēm.
Lai daļas vērtība būtu negatīva, tās skaitītājam un saucējam jābūt ar pretējām zīmēm.
$\begin{array}{l}\frac{f\left(x\right)}{g\left(x\right)}>0\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}⇔\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left\{\begin{array}{l}f\left(x\right)>0\\ g\left(x\right)>0\end{array}\right\\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\mathit{vai}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left\{\begin{array}{l}f\left(x\right)<0\\ g\left(x\right)<0\end{array}\right\\\ \\ \frac{f\left(x\right)}{g\left(x\right)}<0\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}⇔\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left\{\begin{array}{l}f\left(x\right)>0\\ g\left(x\right)<0\end{array}\right\\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\mathit{vai}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left\{\begin{array}{l}f\left(x\right)<0\\ g\left(x\right)>0\end{array}\right\\\ \\ \frac{f\left(x\right)}{g\left(x\right)}\ge 0\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}⇔\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left\{\begin{array}{l}f\left(x\right)\ge 0\\ g\left(x\right)>0\end{array}\right\\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\mathit{vai}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left\{\begin{array}{l}f\left(x\right)\le 0\\ g\left(x\right)<0\end{array}\right\\\ \\ \frac{f\left(x\right)}{g\left(x\right)}\le 0\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}⇔\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left\{\begin{array}{l}f\left(x\right)\ge 0\\ g\left(x\right)<0\end{array}\right\\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\mathit{vai}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left\{\begin{array}{l}f\left(x\right)\le 0\\ g\left(x\right)>0\end{array}\right\\end{array}$
Svarīgi!
Ievēro, daļveida nevienādībai saucējs nedrīkst būt vienāds ar nulli, tāpēc lieto tikai stingro nevienādības zīmi (< vai >).
Piemērs:
$\begin{array}{l}\frac{x+2}{x-3}\ge 0\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\\ \left\{\begin{array}{l}x+2\ge 0\\ x-3>0,\phantom{\rule{0.147em}{0ex}}\mathit{jo}\phantom{\rule{0.147em}{0ex}}x+3\ne 0\end{array}\right\\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\mathit{vai}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left\{\begin{array}{l}x+2\le 0\\ x-3<0\end{array}\right\\\ \\ \left\{\begin{array}{l}x\ge -2\\ x>3\end{array}\right\\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left(1\right)\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\mathit{vai}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\left\{\begin{array}{l}x\le -2\\ x<3\end{array}\right\\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.441em}{0ex}}\left(2\right)\end{array}$
Nevienādību sistēmu atrisinājumu kopas attēlo uz koordinātu taisnēm.

1)

2)

Atbilde:
$x\in \left(-\mathrm{\infty };-2\right]\cup \left(3;+\mathrm{\infty }\right)$

Atsauce:
Matemātika 11. klasei/Evija Slokenberga,Inga France,Ilze France. -Rīga : Lielvārds, 2010. – 320 lpp. :il. – izmantotā literatūra: 14.-15. lpp.