Câu 4: Không có nghĩa khi x-3=0

=>x=3

Câu 5:

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

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

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

ĐKXĐ: \(x\ne3.\)

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

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

\(2,=x^2-3^2=\left(x-3\right)\left(x+3\right)\\ 3,=\left(x+y-x+y\right)\left(x+y+x-y\right)\\ =2y\cdot2x=4xy\)

x^2-9=x^2-3^2=(x+3)(x-3)
(x+y)^2-(x-y)^2=(x+y+x-y)(x+y-x-y)=2x

x²-3²=(x-3)(x+3)

(x+y)²-(x-y)²=(x+y-x+y)(x+y+x-y)=2y⋅2x=4xy

Bài làm 

\(A=\frac{2x+6}{\left(x-3\right)\left(x-2\right)}=\frac{2\left(x+3\right)}{\left(x-3\right)\left(x-2\right)}\)

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

\(A=\frac{2x+6}{\left(x-3\right)\left(x-2\right)}=\frac{2\left(x+3\right)}{\left(x-3\right)\left(x-2\right)}\)

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

\(A=\dfrac{3}{x+3}+\dfrac{1}{x-3}+\dfrac{18}{x^2-9}\)

\(a,\) Điều kiện xác định: \(\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\\x^2-9\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne3\end{matrix}\right.\)

\(b,A=\dfrac{3}{x+3}+\dfrac{1}{x-3}+\dfrac{18}{x^2-9}\)

\(=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}+\dfrac{18}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{4x+12}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{4}{x-3}\)

\(c,x=1\Rightarrow A=\dfrac{4}{1-3}=-2\)