Question

a la a 2 a - b la a 2 a =36 a-b =9 Calculati a + b 3)A=(x€Z | 7 supra 2x+1 €Z) 4)Fie a=(x +3) totul la a 2 a - 2(x-3)(x+3)+(x-3) totul la a 2 a. Arătati ca a este un număr par, orice x€R

Answer 1

$a^2 -b^2 = 36\\\\ a-b = 9\\\\ a^2 - b^2 = (a+b)(a-b) = 9(a+b) = 36\implies a+b = \frac{36}{9} = 4$

$A = \Bigg\{ x\in\mathbb{Z} | \frac{7}{2x+1} \in \mathbb{Z} \Bigg\}\\\\\frac{7}{2x+1} \in \mathbb{Z} \iff 2x+1 \in D_7\\\\ 2x+1 = -7\iff 2x = -8\iff x = -4\\\\ 2x+1 = -1 \iff 2x = -2\iff x = -1\\\\ 2x+1 = 1\iff 2x = 0\iff x = 0\\\\ 2x+1 = 7 \iff 2x = 6\iff x = 3\\\\ A = \Bigg\{-4,-1,0,3\Bigg\}$

$a = (x+3)^2 - 2(x-3)(x+3) + (x-3)^2 = {\Big((x+3)-(x-3)\Big)}^2 = (x+3-x+3)^2 = 6^2 = 36,\forall x \in \mathbb{R}\implies a \in M_2, \forall x \in \mathbb{R}$

Answer 2

Răspuns:

Explicație pas cu pas:

a) a²-b²=36  si a-b=9

a²-b²=(a-b)(a+b)   ⇒a+b=(a²-b²)/(a-b)=36/9=4

A={x∈Z, 7/(2x+1) ∈Z}

2x+1 trebuie sa fie divizor al lui 7: -7,-1,1,7

2x+1=-7     2x=-8                x=-4

2x+1=-1     2x=-2                 x=-1

2x+1=1       2x=0                 x=0

2x+1=7       2x=6                 x=3

A={-4,-1,0,3}

c)a=(x+3)²-2(x+3)(x-3)+(x-3)²=[(x+3)-(x-3)]²=(x+3-x+3)²=6²=36⇒

a este numar intreg natural si par! oricare ar fi x∈R