`a,`

\(x^2-3x\ne0\)

`<=>x(x-3)`\(\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x-3\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne3\end{matrix}\right.\)

`b,`

đặt `A=(x^2-6x+9)/(x^2-3x)`

`A= ((x-3)^2)/(x(x-3))`

`A= (x-3)/x`

`c, `

để `x=5`

`=> A= (x -3)/x=(5-3)/5= 2/5`

a/ ĐKXĐ: \(x^2-3x\ne0\) \(\Leftrightarrow\) x\(\ne\)0,x\(\ne\)3

b/ \(\dfrac{x^2-6x+9}{x^2-3x}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)

c/ x= 5 => \(\dfrac{x-3}{x}=\dfrac{5-3}{5}=\dfrac{2}{5}\)

\(\dfrac{2xy-x^2}{3x^3-6x^2y}\\ =\dfrac{x\left(2y-x\right)}{3x^2\left(x-2y\right)}\\ =\dfrac{x\left(2y-x\right)}{-3x^2\left(2y-x\right)}\\ =\dfrac{1}{-3x}\)

\(\dfrac{2xy-x^2}{3x^3-6x^2y}=\dfrac{-\left(x^2-2xy\right)}{3x^3-6x^2y}\)

\(=\dfrac{-x\left(x-2y\right)}{3x^2\left(x-2y\right)}=\dfrac{-1}{3x}\)

=(3x^2+x^2)/(6x-2x)/(17+5)

=3x^4 / 4x / 22

a: \(=\dfrac{3\left(x-2\right)}{\left(x-2\right)^3}=\dfrac{3}{\left(x-2\right)^2}\)

b: \(=\dfrac{x^2\left(x+2\right)}{\left(x+2\right)^3}=\dfrac{x^2}{\left(x+2\right)^2}\)

\(a,\frac{x^2-8x+15}{x^2-6x+9}\)

\(=\frac{\left(x-4\right)^2-1}{\left(x-3\right)^2}\)

\(=\frac{\left(x-3\right)\left(x-5\right)}{\left(x-3\right)^2}\)

\(=\frac{x-5}{x-3}\)

b) \(\frac{2x^2+3x-2}{x^2+x-2}\)

\(=\frac{2x^2-4x+x-2}{x^2+2x-x-2}\)

\(=\frac{2x\left(x-2\right)+\left(x-2\right)}{x\left(x+2\right)-\left(x+2\right)}\)

\(=\frac{\left(2x+2\right)\left(x-2\right)}{\left(x-1\right)\left(x+2\right)}\)

b)\(\frac{2x^2+3x-2}{x^2+x-2}\)

\(=\frac{\left(2x-1\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}\)

\(=\frac{2x-1}{x-1}\)

Bài 1: 

Để B nguyên thì \(3x+1⋮x-1\)

\(\Leftrightarrow x-1\inƯ\left(4\right)\)

\(\Leftrightarrow x-1\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(x\in\left\{2;0;3;-1;5;-3\right\}\)

Bài 2: 

a: Ta có: \(P=\dfrac{x^2-9}{x^2-6x+9}\)

\(=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2}\)

\(=\dfrac{x+3}{x-3}\)

b: Để P nguyên thì \(x+3⋮x-3\)

\(\Leftrightarrow x-3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

hay \(x\in\left\{4;2;5;1;6;0;9;-3\right\}\)

\(C=\frac{x^4+2x^2+3x^3+6x-2}{x^2+2}\)

\(C=\frac{x^2.\left(x^2+2\right)+3x.\left(x^2+2\right)-2}{x^2+2}\)

\(C=\frac{\left(x^2+3x\right).\left(x^2+2\right)-2}{x^2+2}=\frac{x^2+3x-2}{x^2+2}\)

\(P=\dfrac{3x^2+6x+3}{x+1}\)

\(a,\) Điều kiện xác định: \(x+1\ne0\Leftrightarrow x\ne-1\)

\(b,P=\dfrac{3x^2+6x+3}{x+1}=\dfrac{3\left(x^2+2x+1\right)}{x+1}=\dfrac{3\left(x+1\right)^2}{x+1}=3\left(x+1\right)=3x+3\)

\(c,x=1\Rightarrow P=3.1+3=6\)