biểu thức B đâu rồi bạn

a) \(\left(\frac{x+3}{x-2}+\frac{x+2}{3-x}+\frac{x+2}{x^2-5x+6}\right):\left(\frac{1-x}{x+1}\right)\)

= \(\left(\frac{x+3}{x-2}-\frac{x+2}{x-3}+\frac{x+2}{x^2-2x-3x+6}\right):\left(\frac{1-x}{x+1}\right)\)

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

= \(\left(\frac{x^2-9-x^2+4+x+2}{\left(x-2\right)\left(x-3\right)}\right).\frac{x+1}{1-x}\)

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

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

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

b) Để A >1 \(\Leftrightarrow\frac{x+1}{\left(x-2\right)\left(1-x\right)}>1\)

\(\Leftrightarrow\frac{-\left(1-x\right)\left(3-x\right)}{\left(x-2\right)\left(1-x\right)}\)

\(\Leftrightarrow\frac{x-3}{x-2}>0\)

\(\Rightarrow\orbr{\begin{cases}x-3\ge0\\x-2>0\end{cases}\Leftrightarrow\orbr{\begin{cases}x\ge3\\x>2\end{cases}\Leftrightarrow}x\ge3}\)

\(\Rightarrow\orbr{\begin{cases}x-3< 0\\x-2< 0\end{cases}\Leftrightarrow\orbr{\begin{cases}x< 3\\x< 2\end{cases}\Leftrightarrow}x< 2}\)

Vậy ...

\(\dfrac{x-2+5}{x-2}=1+\dfrac{5}{x-2}\Rightarrow x-2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

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

Muốn giá trị trên thuộc Z =>$\frac{5}{x-2}$ thuộc Z

=>x-2 thuộc Ư(5)

Ta có bảng ( điều kiện:x khác 2 và x thuộc Z 0

Vậy x thuộc 7;3;-3;1

Để \(\dfrac{x+3}{x-2}\) nhận giá trị nguyên thì:

\(\begin{matrix}x+3⋮x-2\\x-2⋮x-2\end{matrix}\)    \(\Leftrightarrow x+3-x+2⋮x-2\)   \(\Leftrightarrow5⋮x-2\)

=> \(x-2\inƯ\left(5\right)=\left\{-1;1;-5;5\right\}\)

Có bảng sau :

Vậy...

\(E=\frac{3-x}{x-1}=\frac{1-x+2}{x-1}=-1+\frac{2}{x-1}\)

E \(\inℤ\Leftrightarrow x+1\inƯ\left(2\right)=\left\{1;-1:2;-2\right\}\)

=> \(x\in\left\{2;0;3;-1\right\}\)

Vậy  \(x\in\left\{2;0;3;-1\right\}\)là giá trị cần tìm

Ta có: \(E=\frac{3-x}{x-1}=\frac{-\left(x-3\right)}{x-1}=\frac{-\left(x-1-2\right)}{x-1}=\frac{-\left(x-1\right)+2}{x-1}=\frac{2}{x-1}-1\)

Để E có giá trị nguyên thì \(\frac{2}{x-1}-1\) có giá trị nguyên

\(\Rightarrow\frac{2}{x-1}\) có giá trị nguyên

\(\Rightarrow x-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

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

1/ Ta có \(\frac{1}{3}< \frac{9}{x}< \frac{1}{2}\)

\(\Rightarrow\frac{9}{27}< \frac{9}{x}< \frac{9}{18}\)

\(\Rightarrow27>x>18\)

Vì \(x\in Z\Rightarrow x\in\left\{19,20,...,26\right\}\)

Vậy....