Question

poza dau coraona urgentt !​

Answer 1

49

a) x+1∈ divizori lui -4⇒x+1∈{1,2,4,-1,-2,-4}⇒x∈{0,1,3,-2,-3,-5}

b)2x+1 ∈ divizori lui -7⇒2x+1∈{1,7,-1,-7}⇒x ∈ {0,3,-1,-4}

c)$\frac{2x+1}{x+2}=\frac{2x+4-3}{x+2}=\frac{2(x+2)}{x+2}+\frac{3}{x+2}=2+ \frac{3}{x+2}$

deoarece 2∈Z ⇒$\frac{3}{x+2}$∈Z⇒x+2 ∈ divizori lui 3 ⇒x ∈ {-1,1,-3,-5}

1

a) -2+3=1 ⇒admite solutia -2

b) -2-4=-6=> nu admite solutia -2

c)4+1=5 => admite solutia -2

d) -3=-3 => admite solutia -2

2

a) x-3=-1 <=> x=-1+3=2

b) 2x+7=3 <=> 2x=3-7 <=> 2x=-4 => x=-2

c)-3x+1=-5 <=> -3x=-6 <=> x=2

d) 2-2x=4 <=> -2x=4-2 <=> -2x=2 <=> x=-1