Đáp án:

`a)` `A= (x+1)/(x-3)`

`b)` `x=5` hoặc `x=2`

Giải thích các bước giải:

`a)`

`A= ((x^2-3)/(x^2-9)+1/(x-3)) : x/(x+3)` `(x \ne 0; x \ne +-3)`

`= [(x^2-3)/((x-3)(x+3))+(x+3)/((x-3)(x+3))]. (x+3)/x`

`= [(x^2-3+x+3)/((x-3)(x+3))]. (x+3)/x`

`= (x^2+x)/((x-3)(x+3)). (x+3)/x`

`= (x(x+1))/((x-3)(x+3)). (x+3)/x`

`= (x+1)/(x-3)`

Vậy `A= (x+1)/(x-3)`

`b)`

Theo bài ra ta có:

`|A| = 3`

`<=> |(x+1)/(x-3)| = 3`

`<=>` \(\left[ \begin{array}{l}\dfrac{x+1}{x-3}=3\\\dfrac{x+1}{x-3}=-3\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}3(x-3) = x+1\\-3(x-3) = x+1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}3x-9=x+1\\-3x+9=x+1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}3x-x=9+1\\-3x-x=-9+1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}2x=10\\-4x=-8\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=5\\x=2\end{array} \right.\) `(` Thỏa mãn `)`

Vậy để `|A| = 3` thì `x=5` hoặc `x=2`

 

Đáp án+Giải thích các bước giải:

`a)A=((x^2 -3)/(x^2 -9) +1/(x-3)) :x/(x+3)``(đk:xne0;xne+-3)`

`A=[(x^2 -3+x+3)/((x-3)(x+3))] .(x+3)/x`

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

`A=(x(x+1))/((x-3)(x+3)) .(x+3)/x`

`A=(x+1)/(x-3)`

Vậy `A=(x+1)/(x-3)``(xne0;xne+-3)`

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`b)|A|=3``(đk:xne0;xne+-3)`

`<=>|(x+1)/(x-3)| =3`

`TH1:(x+1)/(x-3) =3`

`<=>3(x-3)=x+1`

`<=>3x-9=x+1`

`<=>2x=10`

`<=>x=5`$\text{(tm)}$

`TH2:(x+1)/(x-3) =-3`

`<=>-3(x-3)=x+1`

`<=>-3x+9=x+1`

`<=>-4x=-8`

`<=>x=2`$\text{(tm)}$

Vậy với `x=5` hoặc `x=2` thì `|A|=3`