a: \(A=\dfrac{x^2-5x+6-x^2+x+2x^2-6}{x\left(x-3\right)}=\dfrac{2x^2-4x}{x\left(x-3\right)}=\dfrac{2x}{x-3}\)

a) ĐKXĐ: x∉{2;-2}

b) Ta có: \(A=\dfrac{x}{x-2}+\dfrac{2-x}{x+2}+\dfrac{12-10x}{x^2-4}\)

\(=\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}+\dfrac{12-10x}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x^2+2x-x^2+4x-4+12-10x}{\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{-4x+8}{\left(x+2\right)\left(x-2\right)}\)

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

\(=\dfrac{-4}{x+2}\)

c) Để \(A=\dfrac{2}{3}\) thì \(\dfrac{-4}{x+2}=\dfrac{2}{3}\)

\(\Leftrightarrow x+2=\dfrac{-4\cdot3}{2}=-\dfrac{12}{2}=-6\)

hay x=-6-2=-8(nhận)

Vậy: Để \(A=\dfrac{2}{3}\) thì x=-8

d) Để A nguyên thì \(-4⋮x+2\)

\(\Leftrightarrow x+2\inƯ\left(-4\right)\)

\(\Leftrightarrow x+2\in\left\{1;-1;2;-2;4;-4\right\}\)

\(\Leftrightarrow x\in\left\{-1;-3;0;-4;2;-6\right\}\)(nhận)

Vậy: Để A nguyên thì \(x\in\left\{-1;-3;0;-4;2;-6\right\}\)

a, ĐKXĐ:\(\left\{{}\begin{matrix}x+3\ne0\\x^2+x-6\ne0\\2-x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\x^2+x-6\ne0\\x\ne2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)

b, \(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)

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

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

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

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

\(=\dfrac{x-4}{x-2}\)

 \(c,A=\dfrac{-3}{4}\\ \Leftrightarrow\dfrac{x-4}{x-2}=\dfrac{-3}{4}\\ \Leftrightarrow4\left(x-4\right)=-3\left(x-2\right)\\ \Leftrightarrow4x-16x=-3x+6\\ \Leftrightarrow4x-16x+3x-6=0\\ \Leftrightarrow7x-22=0\\ \Leftrightarrow x=\dfrac{22}{7}\)

d, \(A=\dfrac{x-4}{x-2}=\dfrac{x-2-2}{x-2}=1-\dfrac{2}{x-2}\)

Để \(A\in Z\Rightarrow\dfrac{2}{x-2}\in Z\Rightarrow x-2\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Ta có bảng:
 

Vậy \(x\in\left\{0;1;3;4\right\}\)

a)x khác -3 và x khác 2 =)

a: ĐXKĐ: \(x\notin\left\{-3;2\right\}\)

b: \(A=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{1}{x-2}\)

\(=\dfrac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}=\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}=\dfrac{x-4}{x-2}\)

c: Để A=-3/4 thì x-4/x-2=-3/4

=>4x-16=-3x+6

=>7x=22

hay x=22/7

a/

ĐKXĐ: \(x\ne\left\{-1;0;1\right\}\)

b.

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

c.

\(A=2\Rightarrow\dfrac{x+1}{x-1}=2\)

\(\Rightarrow x+1=2x-2\)

\(\Rightarrow x=3\) (thỏa mãn)

d.

\(A=\dfrac{x+1}{x-1}=\dfrac{x-1+2}{x-1}=1+\dfrac{2}{x-1}\)

\(A\) nguyên \(\Leftrightarrow\dfrac{2}{x-1}\) nguyên

\(\Rightarrow x-1=Ư\left(2\right)\)

\(\Rightarrow\left[{}\begin{matrix}x-1=-2\\x-1=-1\\x-1=1\\x-1=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\left(ktm\right)\\x=0\left(ktm\right)\\x=2\left(tm\right)\\x=3\left(tm\right)\end{matrix}\right.\)

Vậy \(x=\left\{2;3\right\}\) thì A nguyên

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{4;1\right\}\end{matrix}\right.\)

Ta có: \(A=\dfrac{x-4\sqrt{x}+3-\left(2x-4\sqrt{x}-\sqrt{x}+2\right)+x+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{2x-4\sqrt{x}+5-2x+5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

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

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{1;4\right\}\end{matrix}\right.\)

\(A=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{x-3\sqrt{x}+2}\)

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

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

\(=\dfrac{x-4\sqrt{x}+3-2x+5\sqrt{x}-2+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\dfrac{1}{\sqrt{x}-2}\)

b: Để A>2 thì A-2>0

=>\(\dfrac{1-2\left(\sqrt{x}-2\right)}{\sqrt{x}-2}>0\)

=>\(\dfrac{5-2\sqrt{x}}{\sqrt{x}-2}>0\)

=>\(\dfrac{2\sqrt{x}-5}{\sqrt{x}-2}< 0\)

TH1: \(\left\{{}\begin{matrix}2\sqrt{x}-5>0\\\sqrt{x}-2< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}>\dfrac{5}{2}\\\sqrt{x}< 2\end{matrix}\right.\)

=>\(x\in\varnothing\)

TH2: \(\left\{{}\begin{matrix}2\sqrt{x}-5< 0\\\sqrt{x}-2>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}< \dfrac{5}{2}\\\sqrt{x}>2\end{matrix}\right.\)

=>\(2< \sqrt{x}< \dfrac{5}{2}\)

=>4<x<25/4

c: Để A là số nguyên thì \(1⋮\sqrt{x}-2\)

=>\(\sqrt{x}-2\in\left\{1;-1\right\}\)

=>\(\sqrt{x}\in\left\{3;1\right\}\)

=>\(x\in\left\{1;9\right\}\)

kết hợp ĐKXĐ, ta được: x=9

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

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

b: Để P<0 thì x+1>0

hay x>-1

c: Để Q=(-2x)/(x+1) là số nguyên thì \(-2x-2+2⋮x+1\)

\(\Leftrightarrow x+1\in\left\{1;-1;2;-2\right\}\)

hay \(x\in\left\{0;-2;-3\right\}\)