$x^2+2x-7⋮x+2$

=> x(x + 2 ) - 7 ⋮ x + 2

=> 7 ⋮ x + 2

=> x + 2 ∈ {1 ; 7 ; -1 ; - 7}

=> x ∈ {-1;5;-3;-9}

=x (x+2) -7  ⋮  x +2 ;=7 ⋮ x+2; = x+2  ∈ \(x^2+2x-7⋮x+2;=x\left(x+2\right)-7⋮x+2;=7⋮x+2;=x+2\in\left\{1;7;-1;-7\right\};=x\in\left\{-1;5;-3;-9\right\}\)

=x (x+2) -7  ⋮  x +2 ;=7 ⋮ x+2; = x+2  ∈ x^2+2x-7⋮x+2;=x\left(x+2\right)-7⋮x+2;=7⋮x+2;=x+2\in\left\{1;7;-1;-7\right\};=x\in\left\{-1;5;-3;-9\right\}x2+2x−7⋮x+2;=x(x+2)−7⋮x+2;=7⋮x+2;=x+2∈{1;7;−1;−7};=x∈{−1;5;−3;−9}

a: =>3x-3+5 chia hết cho x-1

=>x-1 thuộc {1;-1;5;-5}

=>x thuộc {2;0;6;-4}

b: =>x(x+2)-7 chia hết cho x+2

=>x+2 thuộc {1;-1;7;-7}

=>x thuộc {-1;-3;5;-9}

\(3x+2⋮x-1\)

\(\Leftrightarrow3\left(x-1\right)+5⋮x-1\)

\(\Leftrightarrow5⋮x-1\)

\(\Leftrightarrow\left(x-1\right)\inƯ\left(5\right)\)

\(\Leftrightarrow\left(x-1\right)\in\left\{\pm1;\pm5\right\}\)

\(\Leftrightarrow x\in\left\{-4;0;2;6\right\}\)

Vậy để \(3x+2⋮x-1\) thì \(x\in\left\{-4;0;2;6\right\}\)

b) \(x^2+2x-7⋮x+2\)

\(\Leftrightarrow x\left(x+2\right)-7⋮x+2\)

\(\Leftrightarrow7⋮x+2\)

\(\Leftrightarrow\left(x+2\right)\inƯ\left(7\right)\)

\(\Leftrightarrow\left(x+2\right)\in\left\{\pm1;\pm7\right\}\)

\(\Leftrightarrow x\in\left\{-9;-3;-1;5\right\}\)

Vậy để \(x^2+2x-7⋮x+2\) thì \(x\in\left\{-9;-3;-1;5\right\}\)

Bạn ghi lại đề đi bạn, khó hiểu quá!

ồ cuk khó nhỉ

Nếu các bn thích thì ...........

cứ cho NTN này nhé !

\(a,x-5⋮x+2\)

\(\Rightarrow x+2-7⋮x+2\)

\(\Rightarrow x+2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

x + 2 = 1=> x = -1

x + 2 = -1 => x = -3

.... tương tự nhé ~ 

\(2x+3⋮x-5\)

\(\Rightarrow2x-10+7⋮x-5\)

\(\Rightarrow2\left(x-5\right)+7⋮x-5\)

\(\Rightarrow x-5\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

x - 5 = 1 => x = 6 

.... 

lớp 6 rồi nha chú mày