Question

Fie numere naturale nenule x+a²+a,y=a-1 si z=a²-1. Calculand x·y supra z se obtine

Answer 1

$\displaystyle{ x = a^{2} + a }$

$\displaystyle{ y = a - 1 }$

$\displaystyle{ z = a^{2} - 1 }$

$\displaystyle{ \frac{x\cdot y}{z} = \frac{(a^{2}+a) \cdot (a - 1)}{a^{2}-1} }$

$\displaystyle{ \frac{x \cdot y }{z} = \frac{(a^{2} + a) \cdot (a - 1)}{(a-1) \cdot (a+1)} }$

$\displaystyle{ \frac{x \cdot y}{z} = \frac{a^{2}+a}{a+1} = \frac{a \cdot (a + 1)}{a+1} }$

$\displaystyle{ \frac{x \cdot y}{z} = a }$

Answer 2

Răspuns:

Explicație pas cu pas:

x=a²+a=a(a+1)      

y=a-1

z=a²-1=(a-1)(a+1)

xy/z=a(a+1)(a-1)/(a+1)(a-1)=a