a) đk: x khác 0;2;-2;3

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

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

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

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

= \(\dfrac{4x^2+8x}{\left(2-x\right)\left(2+x\right)}.\dfrac{x\left(2-x\right)}{x-3}\)

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

b) Có \(\left|x-5\right|=2\)

<=> \(\left[{}\begin{matrix}x-5=2< =>x=7\left(Tm\right)\\x-5=-2< =>x=3\left(L\right)\end{matrix}\right.\)

Thay x = 7 vào A, ta có:

\(A=\dfrac{4.7^2}{7-3}=49\)

c) A = \(\dfrac{4x^2}{x-3}⋮4\left(\forall x\right)\)

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: \(A=\dfrac{x^2+1}{x}+\dfrac{x^3-1}{x^2-x}+\dfrac{x^4-x^3+x-1}{x-x^3}\)

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

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

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

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

b: \(x^2+x=12\)

=>\(x^2+x-12=0\)

=>(x+4)(x-3)=0

=>\(\left[{}\begin{matrix}x+4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(nhận\right)\\x=-4\left(loại\right)\end{matrix}\right.\)

Thay x=3 vào A, ta được:

\(A=\dfrac{\left(3+1\right)^2}{3}=\dfrac{16}{3}\)

Khi x=-4 thì \(A=\dfrac{\left(-4+1\right)^2}{-4}=\dfrac{9}{-4}=-\dfrac{9}{4}\)

c: \(A-4=\dfrac{\left(x+1\right)^2}{x}-4\)

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

\(=\dfrac{x^2+2x+1-4x}{x}=\dfrac{x^2-2x+1}{x}=\dfrac{\left(x-1\right)^2}{x}\)>0 với mọi x>0

=>A>4

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

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

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

b: |x-3|=2

=>x-3=2 hoặc x-3=-2

=>x=5(nhận) hoặc x=1(loại)

Khi x=5 thì \(E=\dfrac{5^2}{5-1}=\dfrac{25}{4}\)

c: Để E=1/2 thì \(\dfrac{x^2}{x-1}=\dfrac{1}{2}\)

\(\Leftrightarrow2x^2-x+1=0\)

hay \(x\in\varnothing\)

f) \(A=\dfrac{x^2}{x-1}=\dfrac{x^2-x+x-1+1}{x-1}=\dfrac{x\left(x-1\right)+x-1+1}{x-1}=x+1+\dfrac{1}{x-1}=x-1+\dfrac{1}{x-1}+2\ge2\sqrt{\left(x-1\right).\dfrac{1}{x-1}}+2=4\)\(A=4\Leftrightarrow x=2\)

-Vậy \(A_{min}=4\)

a: \(A=\dfrac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{6\left(x+2\right)}\)

\(=\dfrac{-6}{\left(x+2\right)}\cdot\dfrac{-\left(x+1\right)}{6\left(x+2\right)}=\dfrac{\left(x+1\right)}{\left(x+2\right)^2}\)

b: A>0

=>x+1>0

=>x>-1

c: x^2+3x+2=0

=>(x+1)(x+2)=0

=>x=-2(loại) hoặc x=-1(loại)

Do đó: Khi x^2+3x+2=0 thì A ko có giá trị

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: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{x-9}\)

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

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