Answer:

a. \(ĐKXĐ:x^2-9\ne0\Rightarrow x^2\ne9\Rightarrow x\ne\pm3\)

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

c. \(A=7\)

\(\Rightarrow\frac{x-3}{x+3}=7\)

\(\Rightarrow x-3=7.\left(x+3\right)\)

\(\Rightarrow x-3=7x+21\)

\(\Rightarrow x-3-7x-21=0\)

\(\Rightarrow-6x-24=0\)

\(\Rightarrow x=-4\)

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

a: ĐKXĐ: x<>-3

b: =x+3

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

a) ĐKXĐ:

\(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)

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

c) Thay \(x=-2\) vào A, ta có:

\(A=\dfrac{1}{-2+1}=-1\)

Vậy khi x = -2 thì A = -1

a) ĐKXĐ:   \(x\ne\pm1\)

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

c) Khi x = - 2 

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

Vậy khi x = - 2 thì biểu thức có giá trị bằng - 1

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

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

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

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

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

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

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

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

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

a) Phân thức xác định khi: \(\Leftrightarrow x-3\ne3\Leftrightarrow x\ne3\)

ĐKXĐ: \(x\ne3\)

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

c) Thay x = -4 vào phân thức đã thu gọn, ta có:

 \(A=\frac{2.\left(-4\right)}{\left(-4\right)-3}=\frac{8}{7}\)

Vậy: tại x = -4 là \(\frac{8}{7}\)

a) \(x^2-9=\left(x-3\right)\left(x+3\right)\)

Phân thức xác định khi: \(\left(x-3\right)\left(x+3\right)\ne0\)

\(\Leftrightarrow\hept{\begin{cases}x-3=0\\x+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\x=-3\end{cases}}\Leftrightarrow x\ne\pm3\)

ĐKXĐ: \(x\ne\pm3\)

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

c) \(A=\frac{2.\left(-4\right)}{\left(-4\right)-3}=\frac{8}{7}\)

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

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

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

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