368/MR00. zadatak / Matematika - Zbirka rešenih zadataka

MR-368 / 12. zadatak

Odrediti istinitosne vrednosti iskaza:

$p \equiv \frac{3 + 2 x}{3} - \frac{2 - 3 x}{4} = \frac{29}{24}; x = 0, 5$

$q \equiv 3 (5 - y) - 2 (y - 1) = 1 + 3 y; y = 2$ pa zatim odrediti istinitosne vrednosti sledećih iskaza:

a) $(p \lor q) \land p$

b) $(p \land q) \lor p$

$p \equiv \frac{3 + 2 x}{3} - \frac{2 - 3 x}{4} = \frac{29}{24}; x = 0, 5$

$p \equiv \frac{3 + 2 \cdot 0, 5}{3} - \frac{2 - 3 \cdot 0, 5}{4} = \frac{29}{4}$

p \equiv \frac{3 + 1}{3} - \frac{2 - 1, 5}{4} = \frac{29}{4}

p \equiv \frac{4}{3} - \frac{0, 5}{4} = \frac{29}{4}

p \equiv \frac{4}{3} - \frac{1}{4} \cdot \frac{1}{2} = \frac{29}{4}

p \equiv \frac{4}{3} - \frac{1}{8} = \frac{29}{4}

$p \equiv \frac{4}{3} - \frac{1}{8} = \frac{4 \cdot 8 - 3 \cdot 1}{24} = \frac{32 - 3}{24} = \frac{29}{24} = \frac{29}{24}$

p \equiv \frac{29}{24} = \frac{29}{24} \Rightarrow p = ⊤

$q \equiv 3 (5 - y) - 2 (y - 1) = 1 + 3 y; y = 2$

q = 3 (5 - 2) - 2 (2 - 1) = 1 + 3 \cdot 3

q = 3 \cdot 3 - 2 \cdot 1 = 1 + 3 \cdot 2

q = 9 - 2 = 1 + 6

q = 7 = 7 \Rightarrow q = ⊤

$a) (p \lor q) \land p = (⊤ \lor ⊤) \land ⊤ = ⊤ \land ⊤ = ⊤$

(p \lor q) \land p = ⊤

$b) (p \land q) \lor p = (⊤ \land ⊤) \lor ⊤ = ⊤ \lor ⊤ = ⊤$

(p \land q) \lor p = ⊤

$a) (p \land q) \lor p = ⊤ b) (p \lor q) \land p = ⊤$

Zamenite vrednost x u jednačinu, i proverite istinitost iskaza

$a) (p \land q) \lor p = ⊤ b) (p \lor q) \land p = ⊤$

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