tìm x thuộc z để các phép chia sau hết a)2x^2-x/x+1 b)-x^2+2x-5/x-2
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Bài 1:
Ta có: \(5x^3-3x^2+2x+a⋮x+1\)
\(\Leftrightarrow5x^3+5x^2-8x^2-8x+10x+10+a-10⋮x+1\)
\(\Leftrightarrow a-10=0\)
hay a=10

\(a,x-5⋮x+2\)
\(\Rightarrow x+2-7⋮x+2\)
\(\Rightarrow x+2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
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)Ta có : \(x-5⋮x+2=>x-5-\left(x+2\right)⋮x-2=>-7⋮x-2\)
\(=>x-2\inƯ\left(7\right)\left\{-7;-1;1;7\right\}\)
\(=>x\in\left\{-5;1;3;9\right\}\)
b)Ta có : \(2x+1⋮2x-1=>2x+1-\left(2x-1\right)⋮2x-1=>2⋮2x-1\)
\(=>2x-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
\(=>2x\in\left\{-1;0;2;3\right\}\)
\(=>x\in\left\{0;1\right\}\)(vì \(x\in Z\))
c)\(\left(x+5\right)-3\left(x+5\right)+2⋮x+5=>2⋮x+5=>x+5\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
\(=>x\in\left\{-7;-6;-4;-3\right\}\)
d)\(x+1⋮x+2=>x+2-1⋮x+2\)
\(=>1⋮x+2=>x+2\inƯ\left(1\right)=\left\{1;-1\right\}=>x\in\left\{-1;-3\right\}\)

a)Ta có:2x4-2x3+x2+x+a
= 2x3(x-2)+2x2(x-2)+5x(x-2)+11(x-2)+a+22
= (x-2)(2x3+2x2-5x+11)+(a+22)
Để (x-2)(2x3+2x2-5x+11)+(a+22)⋮(x-2) thì a+22=0⇔a=-22
b)Ta có:2x3-3x2+x+a
= 2x2(x+2)-5x(x+2)+11(x+2)+(a-22)
= (x+2)(2x2-5x+11)+(a-22)
Để (x+2)(2x2-5x+11)+(a-22)⋮(x+2) thì a-22=0⇔a=22
a)Ta có: \(2x^2-x⋮x+1\)
\(\Leftrightarrow2x^2+2x-3x-3+3⋮x+1\)
\(\Leftrightarrow x+1\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{0;-2;2;-4\right\}\)
b) Ta có: \(-x^2+2x-5⋮x-2\)
\(\Leftrightarrow-5⋮x-2\)
\(\Leftrightarrow x-2\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{3;1;7;-3\right\}\)