Để \(M=\dfrac{x^2+2x-13}{x-3}\in Z\) thì \(x^2+2x-13⋮x-3\)
\(\Rightarrow\left(x^2-3x\right)+5x-13⋮x-3\)
\(\Rightarrow x\left(x-3\right)+5x-13⋮x-3\)
\(\Rightarrow5x-13⋮x-3\)
\(\Rightarrow\left(5x-15\right)+2⋮x-3\)
\(\Rightarrow5\left(x-3\right)+2⋮x-3\)
\(\Rightarrow2⋮x-3\)
\(\Rightarrow x-3\in U\left(2\right)=\left\{-1;1;-2;2\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}x-3=-1\Rightarrow x=2\\x-3=1\Rightarrow x=4\\x-3=-2\Rightarrow x=1\\x-3=2\Rightarrow x=5\end{matrix}\right.\)
Vậy \(x\in\left\{2;4;1;5\right\}\) thì \(M\in Z\)
a) \(x^2-9x\)
\(=x\left(x-9\right)\)
b) \(3x^2-3xy-5x+5y\)
\(=\left(3x^2-3xy\right)-\left(5x-5y\right)\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
c) \(x^3+5x^2-6x\)
\(=x\left(x^2+5x-6\right)\)
\(=x\left(x^2-x+6x-6\right)\)
\(=x\left[\left(x^2-x\right)+\left(6x-6\right)\right]\)
\(=x\left[x\left(x-1\right)+6\left(x-1\right)\right]\)
\(=x\left(x-1\right)\left(x+6\right)\)
a, =x(2-9)=-7x
b,3x^2-5x-(3xy-5y)=x(3x-5)-(y(3x-5))=(3x-5)(x-y)
d, =x(x^2+5x-6)
a) x2-9x
=x(x-9)
b)3x2-3xy-5x+5y
=(3x2-3xy)-(5x-5y)
=3x(x-y)-5(x-y)
=(x-y)(3x-5)
d)x3+5x2-6x
=x(x2+5x-6)
=x(x2+6x-x-6)
=x[(x2+6x)-(x+6)]
=x[x(x+6)-(x+6)]
=x(x+6)(x-1)
