M = \(\left(\frac{9}{x\left(x^2-9\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)

<=> M = 

sẽ có k cho các bạn làm vừa nhanh vừa đúng ! giúp mk nha ! thank you !!! 

a) Để M là số dương, có 2 TH:

\(TH1:7x-3>0;x+2>0\)

\(\Rightarrow x>\frac{3}{7};x>-2\)

\(TH2:7x-3< 0;x+2< 0\)

\(\Rightarrow x< \frac{3}{7};x< -2\)

b) Để M là số âm có 2 TH:

\(TH1:7x-3< 0;x+2>0\)

\(\Rightarrow x< \frac{3}{7};x>-2\)

\(\Rightarrow-2< x< \frac{3}{7}\)

\(TH2:7x-3>0;x+2< 0\)

\(\Rightarrow x>\frac{3}{7};x< -2\)(vô lí)

Vậy ta chỉ xét được 3 TH

\(\frac{7x-3}{x+2}\Leftrightarrow\frac{7x+14-17}{x+2}\Leftrightarrow\frac{7\left(x+2\right)-17}{x+2}\Leftrightarrow\frac{7\left(x+2\right)}{x+2}-\frac{17}{x+2}\Leftrightarrow7-\frac{17}{x+2}\)

7-17/(x+2) => x khác -2 => 17/(x+2) thuộc Ư(17) ={1,17}

                                      => 17/(x+2) thuộc Ư(17)={-1,-17}

ĐKXĐ: \(x\ne\pm3\)

a

Khi x = 1:

\(A=\dfrac{3.1+2}{1-3}=\dfrac{5}{-2}=-2,5\)

Khi x = 2:

\(A=\dfrac{3.2+2}{2-3}=-8\)

Khi x = \(\dfrac{5}{2}:\)

\(A=\dfrac{3.2,5+2}{2,5-3}=\dfrac{9,5}{-0,5}=-19\)

b

Để A nguyên => \(\dfrac{3x+2}{x-3}\) nguyên

\(\Leftrightarrow3x+2⋮\left(x-3\right)\\3\left(x-3\right)+11⋮\left(x-3\right) \)

Vì \(3\left(x-3\right)⋮\left(x-3\right)\) nên \(11⋮\left(x-3\right)\)

\(\Rightarrow\left(x-3\right)\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\\ \Rightarrow x\left\{4;2;-8;14\right\}\)

c

Để B nguyên => \(\dfrac{x^2+3x-7}{x+3}\) nguyên

\(\Rightarrow x\left(x+3\right)-7⋮\left(x+3\right)\)

\(\Rightarrow-7⋮\left(x+3\right)\\ \Rightarrow x+3\inƯ\left\{\pm1;\pm7\right\}\)

\(\Rightarrow x=\left\{-4;-11;-2;4\right\}\)

d

\(\left\{{}\begin{matrix}A.nguyên.\Leftrightarrow x=\left\{-8;2;4;14\right\}\\B.nguyên\Leftrightarrow x=\left\{-11;-4;-2;4\right\}\end{matrix}\right.\)

=> Để A, B cùng là số nguyên thì x = 4.