(x+3):hết cho (x+1) (2x+7):hết cho (x+1) tìm x

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1. a, Ta có: (x+3)⋮⋮(x+1)⇒(x+1+2)⋮⋮(x+1)⇒(x+1)∈Ư(2)={±1;±2}

   x+1=1⇒x=0

   x+1=-1⇒x=-2

   x+1=2⇒x=1

   x+1=-2⇒x=-3

   Vậy x∈{0;-2;1;-3}
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2. a, Ta có: (x+3)$\vdots$(x+1)⇒(x+1+2)$\vdots$(x+1)⇒(x+1)∈Ư(2)={±1;±2}

   x+1=1⇒x=0

   x+1=-1⇒x=-2

   x+1=2⇒x=1

   x+1=-2⇒x=-3

   Vậy x∈{0;-2;1;-3}

   b, Ta có: (2x+7)$\vdots$(x+1)⇒2(x+1)+5$\vdots$(x+1)⇒(x+1)∈Ư(5)={±1;±5}

   x+1=1⇒x=1

   x+1=-1⇒x=-2

   x+1=5⇒x=4

   x+1=-5⇒x=-6

   Vậy x∈{1;-2;4;-6}

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