a) Ta có: 

Để M = \(\frac{x+3}{2}\)\(\in\)Z <=> \(x+3⋮2\) <=> \(x+3\in\)B(2) = {0; 2; 4; ....}

                                                           <=> \(x\in\){-3; -1; 1; ....}

b) Để N = \(\frac{7}{x-1}\)\(\in\)Z <=> \(7⋮x-1\) <=> \(x-1\in\)Ư(7) = {1; -1; 7; -7}

Lập bảng :

Vậy ...

c) Ta có: P = \(\frac{x-1}{x+1}=\frac{x+1-2}{x+1}=1-\frac{2}{x+1}\)

Để P \(\in\)Z <=> \(2⋮x+1\) <=> \(x+1\in\)Ư(2) = {1; -1; 2; -2}

Lập bảng: 

Vậy ...

để M nguyên thì \(\frac{x+3}{2}\) nguyên 

=> (x+3) \(\in\)Ư(2)={-2:-1:1:2}

lập bảng ra tìm x nha bn ~!!

mấy ý kia tương tự !

a) \(M=\frac{x+3}{2}\in Z\)

=> x+3 chia hết cho 2

=> x+3 thuộc Ư(2)={-1,-2,1,2}

=> x thuộc {-4,-5,-2,-1}

b) \(N=\frac{7}{x-1}\in Z\)

=> x-1 thuộc Ư(7)={-1,-7,1,7}

=> x thuộc {0,-6,2,8}

help me !!!!!!!!!!!!!!

a, Để P xác định <=> \(\hept{\begin{cases}x+3\ne0\\x^2+x-6\ne0\\2-x\ne0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ne-3\\x^2-2x+3x-6\ne\\x\ne2\end{cases}0\Rightarrow\hept{\begin{cases}x\ne-3\\\left(x-2\right)\\x\ne2\end{cases}}}\left(x+3\right)\ne0\)

\(\Leftrightarrow\hept{\begin{cases}x\ne-3\\x\ne2\end{cases}}\)

Rút gọn

\(P=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)

\(=\frac{x+2}{x+3}-\frac{5}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x-2}\)

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

\(=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{x^2-4x+3x-12}{\left(x+3\right)\left(x+2\right)}=\frac{x\left(x-4\right)+3\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{\left(x+3\right)\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)

b,Để \(P=\frac{-3}{4}\)

Thì \(\frac{x-4}{x-2}=\frac{-3}{4}\)

\(\Rightarrow4x-16=-3x+6\)

\(\Rightarrow4x-16-3x+6=0\)

\(\Rightarrow x-10=0\)

\(\Rightarrow x=10\left(t/m\right)\)

Vậy \(P=\frac{-3}{4}\)khi x=10

c,Để \(P\inℤ\Rightarrow x-4⋮x-2\)

mà \(x-4=\left(x-2\right)-2\)

Vì \(x-2⋮\left(x-2\right)\Rightarrow-2⋮\left(x-2\right)\)

\(\Rightarrow x-2\inƯ\left(-2\right)=\left\{\pm1,\pm2\right\}\)

\(\Rightarrow x\in\left\{3,1,4,0\right\}\left(t/m\right)\)

Vậy ......................

d,\(x^2-9=0\)

\(\Rightarrow x^2=9\)

\(\Rightarrow x=\pm3\)

TH1   

Thay x= 3 ta có 

\(P=\frac{3-4}{3-2}\)

\(=\frac{-1}{1}=-1\)

TH2

\(x=-3\)

Vậy \(P=-1\Leftrightarrow x=3\)

e,Để P >0 khi 

\(\orbr{\begin{cases}\hept{\begin{cases}x-4>0\\x-2>0\end{cases}}\\\hept{\begin{cases}x-4< 0\\x-2< 0\end{cases}}\end{cases}}\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x>4\\x>2\end{cases}}\\\hept{\begin{cases}x< 4\\x< 2\end{cases}}\end{cases}}\Rightarrow\orbr{\begin{cases}x>4\\x< 2\end{cases}}\)

Vậy \(P>0\Leftrightarrow\orbr{\begin{cases}x>4\\x< 2\&x\ne-3\end{cases}}\)

a) \(\frac{1-x}{x+4}=\frac{5-4-x}{x+4}=\frac{5}{x+4}-1\inℤ\Leftrightarrow\frac{5}{x+4}\inℤ\)

mà \(x\inℤ\Rightarrow x+4\inƯ\left(5\right)=\left\{-5,-1,1,5\right\}\)

\(\Leftrightarrow x\in\left\{-9,-5,-3,1\right\}\)

b) \(\frac{11-2x}{x-5}=\frac{1+10-2x}{x-5}=\frac{1}{x-5}-2\inℤ\Leftrightarrow\frac{1}{x-5}\inℤ\)

mà \(x\inℤ\Rightarrow x-5\inƯ\left(1\right)=\left\{-1,1\right\}\Leftrightarrow x\in\left\{4,6\right\}\)

c) \(\frac{x+1}{2x+1}\inℤ\Rightarrow\frac{2\left(x+1\right)}{2x+1}=\frac{2x+1+1}{2x+1}=1+\frac{1}{2x+1}\inℤ\Leftrightarrow\frac{1}{2x+1}\inℤ\)

mà \(x\inℤ\Rightarrow2x+1\inƯ\left(1\right)=\left\{-1,1\right\}\Leftrightarrow x\in\left\{-1,0\right\}\).

Thử lại đều thỏa mãn. 

a) x=1/-3/3/6

b)x=?