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

Câu 4: Không có nghĩa khi x-3=0

=>x=3

Câu 5:

\(A=\dfrac{x-3}{\left(x-3\right)\left(x+3\right)}=\dfrac{1}{x+3}\)

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

1. = \(\dfrac{x+y}{x-y}\)
2. = \(\dfrac{x}{x+3}\)

a: ĐKXĐ: x<>-3

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

a) ĐKXĐ:  \(x\ne-3\)

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

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

ĐKXĐ: \(x\ne3.\)

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

b: \(B=\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\)

\(=\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{x+y+z}\)

=x+y-z

c: 
ĐKXĐ: x<>1

\(C=\dfrac{x^2-6x+5}{x^2-2x+1}\)

\(=\dfrac{\left(x-1\right)\left(x-5\right)}{\left(x-1\right)^2}\)

\(=\dfrac{x-5}{x-1}\)