Mathematic Problems

368. példa

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 = ⊤

Helyettesítse be az x érékét a kifelyezésbe, és ellenőrizze annak helyességét

Replace the value of x to in the equation and check its correctness

Zamenite vrednost x u jednačinu, i proverite istinitost iskaza

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

Matematika - Képlettár