Question

Urgent tot exericitul 1 ! Multumesc anticipat!

a). x+1 | x+7
b). x+1 | 3x+11
c). x+2 | 5x+22
d). 2x+5 | 9x+29
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Answer 1

Explicație pas cu pas:

a)

x+1 | x+7 <=> (x+1) | [(x+1)+6]

(x+1) | (x+1) => (x+1) | 6

(x+1) ∈ {1, 2, 3, 6} |-1

=> x ∈ {0, 1, 2, 5}

.

b)

x+1 | 3x+11 <=> (x+1) | [3(x+1) + 8]

(x+1) | 3(x+1) => (x+1) | 8

(x+1) ∈ {1, 2, 4, 8} |-1

=> x ∈ {0, 1, 3, 7}

.

c)

x+2 | 5x+22 <=> (x+2) | [5(x+2) + 12)

(x+2) | 5(x+2) => (x+2) | 12

(x+2) ∈ {1, 2, 3, 4, 6, 12} |-2

=> x ∈ {-1, 0, 1, 2, 4, 10}

x ∈ N => x ∈ {0, 1, 2, 4, 10}

.

d)

2x+5 | 9x+29 <=> (2x+5) | 2×(9x+29)

(2x+5) | (18x+58) <=> (2x+5) | [9×(2x+5)+13]

(2x+5) | 9×(2x+5) => (2x+5) | 13

(2x+5) ∈ {1, 13} |-5

=> 2x ∈ {-4, 8} |:2

=> x ∈ {-2, 4}

x ∈ N => x = 4