- Detalji
- Hitova: 3116
Zadatak 11. Riješi nejednadžbu
\[ \dfrac{3x+1}{x-3}\leq -3 \]
Rješenje: Kao prvo napraviti nlu na desnoj strani
\[ \dfrac{3x+1}{x-3}-3\leq 0 \]
\[ \dfrac{6x-8}{x-3}\leq 0 \]
a) | \(6x-8\leq 0\) | b) | \(6x-8\geq 0\) | |||||
\(x-3>0\) | \( x-3<0\) | |||||||
\(x\leq \frac{4}{3}\) | \(x\geq\dfrac{4}{3}\) | |||||||
\(x>3\) | \(x<3\) | |||||||
\(R_a: x\in \emptyset\) | \(R_b: x\in \left[\frac{4}{3},3\right>\) | |||||||
Rješenje zadatka je unija rješenja sustava pod a) i b):
\[ R=R_a\cup R_b=\left[ \dfrac{4}{3},3\right>\]
- Detalji
- Hitova: 9599
\(\large{\mathbf{1.}\mspace{3mu}Riješi\:\:nejednadžbe:}\)
\(\mspace{50mu}\large{\mathbf{1)}\mspace{15mu}7-4x<0;\mspace{30mu}\mathbf{2)}\mspace{15mu}3x+1\geq x+7;}\)
\(\mspace{50mu}\large{\mathbf{3)}\mspace{15mu}\dfrac{1}{5}x+6<x;\mspace{30mu}\mathbf{4)}\mspace{15mu}x+11\leq 3x-5;}\)
\(\mspace{50mu}\large{\mathbf{5)}\mspace{15mu}3-9x<0;\mspace{30mu}\mathbf{6)}\mspace{15mu}-5-2x>0;}\)
\(\mspace{50mu}\large{\mathbf{7)}\mspace{15mu}-3x+1\leq 2;\mspace{30mu}\mathbf{8)}\mspace{15mu}2x+5\geq -x-7;}\)