tìm x sao cho x^2-2x+7/x^2-2x+3 thuộc Z
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x2+2x−7⋮x+2x2+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 ∈ \(x^2+2x-7⋮x+2;=x\left(x+2\right)-7⋮x+2;=7⋮x+2;=x+2\in\left\{1;7;-1;-7\right\};=x\in\left\{-1;5;-3;-9\right\}\)
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 \(x\in\left\{-9;-3;-1;5\right\}\)
\(a,x-5⋮x+2\)
\(\Rightarrow x+2-7⋮x+2\)
\(\Rightarrow x+2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
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
....
\(A=\dfrac{x^2-2x+7}{x^2-2x+3}=1+\dfrac{4}{x^2-2x+3}=1+\dfrac{4}{\left(x-1\right)^2+2}\)
\(A\in Z\Leftrightarrow\)\(\left[\left(x-1\right)^2+2\right]\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Rightarrow\left(x-1\right)^2+2=2\)
\(\Leftrightarrow x=0\)