cảm ơn bạn nhiều.

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

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

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

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

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

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

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

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

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

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) Điều kiện xác định của \(P\) là: 

\(\left(x+1\right)\left(2x-6\right)\ne0\Leftrightarrow\left\{{}\begin{matrix}x+1\ne0\\2x-6\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\)

b) \(P=\dfrac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}\) (\(x\ne-1,x\ne3\))

\(=\dfrac{3x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{3x}{2\left(x-3\right)}\)

\(P=1\Rightarrow\dfrac{3x}{2\left(x-3\right)}=1\Rightarrow3x=2\left(x-3\right)\Leftrightarrow x=-6\) (thỏa mãn) 

c) \(P>0\Rightarrow\dfrac{3x}{2\left(x-3\right)}>0\Leftrightarrow\dfrac{x}{x-3}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x-3>0\end{matrix}\right.\\\left[{}\begin{matrix}x< 0\\x-3< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>3\\x< 0\end{matrix}\right.\)

Kết hợp với điều kiện xác định ta được để \(P>0\) thì \(x>3\) hoặc \(x< 0,x\ne-1\).

Sửa đề: \(\left\lbrack\frac{4}{x-4}-\frac{4}{x+4}\right\rbrack\cdot\frac{x^2+8x+16}{32}\)

Đặt \(A=\left\lbrack\frac{4}{x-4}-\frac{4}{x+4}\right\rbrack\cdot\frac{x^2+8x+16}{32}\)

a: ĐKXĐ: x∉{4;-4}

b: \(A=\left\lbrack\frac{4}{x-4}-\frac{4}{x+4}\right\rbrack\cdot\frac{x^2+8x+16}{32}\)

\(=\frac{4\left(x+4\right)-4\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}\cdot\frac{\left(x+4\right)^2}{32}\)

\(=\frac{4x+16-4x+16}{x-4}\cdot\frac{x+4}{32}=\frac{32}{x-4}\cdot\frac{x+4}{32}=\frac{x+4}{x-4}\)

\(A=\frac13\)

=>\(\frac{x+4}{x-4}=\frac13\)

=>3(x+4)=x-4

=>3x+12=x-4

=>2x=-16

=>x=-8(nhận)

c: A=1

=>x+4=x-4

=>4=-4(loại)

=>x∈∅

d: Để A nguyên thì x+4⋮x-4

=>x-4+8⋮x-4

=>8⋮x-4

=>x-4∈{1;-1;2;-2;4;-4;8;-8}

=>x∈{5;3;6;2;8;0;12;-4}

Kết hợp ĐKXĐ, ta được: x∈{5;3;6;2;8;0;12}

e: Để A>0 thì \(\frac{x+4}{x-4}>0\)

=>x-4>0 hoặc x+4<0

=>x>4 hoặc 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

còn c và d thôi bạn ơi

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