`a,`để `x` xác định thì

\(3x+9\ne0\)

\(\Leftrightarrow x\ne-3\)

`b,` tại `x=2` thì :

`A=(x^2 + 3)/(3x + 9) =(2^2 +3)/(3.2+9)=(4+3)/(6+9)=7/15`

`=>A=7/15`

Bài 3:

\(C=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)

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

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

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

\(A=\left(\dfrac{2x^2}{x^2-9}+\dfrac{3}{x-3}-\dfrac{x}{x+3}\right).\dfrac{4}{5x+15}\) (1)

a) ĐKXĐ: \(x\ne\pm3\)

b) \(\left(1\right)=\left[\dfrac{2x^2}{\left(x-3\right)\left(x+3\right)}+\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right].\dfrac{4}{5x+15}\)

\(=\dfrac{2x^2+3x+9-x^2+3x}{\left(x-3\right)\left(x+3\right)}.\dfrac{4}{5x+15}\)

\(=\dfrac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}.\dfrac{4}{5\left(x+3\right)}\)

\(=\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}.\dfrac{4}{5\left(x+3\right)}\)

\(=\dfrac{4}{5\left(x-3\right)}\)

c) Thay \(x=19\) vào \(A=\dfrac{4}{5\left(x-3\right)}\) ta có:

\(A=\dfrac{4}{5.\left(19-3\right)}=\dfrac{4}{80}=\dfrac{1}{20}\)

Vậy \(x=19\) thì \(A=\dfrac{1}{20}\)

a) ĐK: \(x\)≠\(+-3\)

b) \(A=\left(\dfrac{2x^2}{x^2-9}+\dfrac{3}{x-3}-\dfrac{x}{x+3}\right).\dfrac{4}{5x+15}\)

\(=\dfrac{2x^2+3\left(x+3\right)-x\left(x-3\right)}{x^2-9}.\dfrac{4}{5\left(x+3\right)}\)

\(=\dfrac{2x^2+3x+9-x^2+3x}{\left(x+3\right)\left(x-3\right)}.\dfrac{4}{5\left(x+3\right)}\)

\(=\dfrac{4\left(x^2+6x+9\right)}{5\left(x+3\right)^2\left(x-3\right)}=\dfrac{4\left(x+3\right)^2}{5\left(x+3\right)^2\left(x-3\right)}=\dfrac{4}{5\left(x-3\right)}=\dfrac{4}{5x-15}\)

c) Tại x=19

⇒ \(A=\dfrac{4}{5.19-15}=\dfrac{4}{80}=\dfrac{1}{20}\)

Vậy ...

=x2+6x+9(x−3)(x+3).45(x+3)=x2+6x+9(x−3)(x+3).45(x+3)

=45(x−3)=45(x−3)

c) Thay  vào A=45.(19−3)=480=120A=45.(19−3)=480=120

Vậy  thì 

a) ĐKXĐ: 

\(x^2-1\ne0\Leftrightarrow x\ne\pm1\)

b) \(A=\dfrac{x^2-2x+1}{x^2-1}\)

\(A=\dfrac{x^2-2\cdot x\cdot1+1^2}{x^2-1^2}\)

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

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

c) Thay x = 3 vào A ta có:

\(A=\dfrac{3-1}{3+1}=\dfrac{2}{4}=\dfrac{1}{2}\)

a) ĐKXĐ: 

\(9x^2-y^2\ne0\Leftrightarrow\left(3x\right)^2-y^2\ne0\Leftrightarrow\left(3x-y\right)\left(3x+y\right)\ne0\)

\(\Leftrightarrow3x\ne\pm y\) 

b) \(B=\dfrac{6x-2y}{9x^2-y^2}\)

\(B=\dfrac{2\cdot3x-2y}{\left(3x\right)^2-y^2}\)

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

\(B=\dfrac{2}{3x+y}\)

Thay x = 1 và \(y=\dfrac{1}{2}\) và B ta có:

\(B=\dfrac{2}{3\cdot1+\dfrac{1}{2}}=\dfrac{2}{3+\dfrac{1}{2}}=\dfrac{2}{\dfrac{7}{2}}=\dfrac{4}{7}\)

a,ĐK:  \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)

b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)

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

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

c, Với x = 4 thỏa mãn ĐKXĐ thì

\(A=\frac{-3}{4-3}=-3\)

d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)

\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)

Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)

\(a,ĐK:x\ne-3;x\ne0;y\ne0\\ b,A=\dfrac{1}{x^2\left(x+3\right)+y^2\left(x+3\right)}=\dfrac{1}{\left(x^2+y^2\right)\left(x+3\right)}\\ x=y=0\Leftrightarrow A\in\varnothing\)

a) ĐKXĐ: 

\(\left\{{}\begin{matrix}x^2-9\ne0\\x+3\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\ne-3\end{matrix}\right.\Leftrightarrow x\ne\pm3\) 

b) \(A=\dfrac{x+15}{x^2-9}-\dfrac{2}{x+3}\)

\(A=\dfrac{x+15}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{x+15-2x+6}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{21-x}{\left(x+3\right)\left(x-3\right)}\)

c) Thay x = - 1 vào A ta có: 

\(A=\dfrac{21-\left(-1\right)}{\left(-1+3\right)\left(-1-3\right)}=\dfrac{21+1}{2\cdot-4}=\dfrac{22}{-8}=-\dfrac{11}{4}\)