Lời giải của bạn Nhật Linh đúng rồi, tuy nhiên cần thêm điều kiện để A có nghĩa: \(x\ne\pm2\)

a,ĐK:  \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)

b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)

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

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

c, Với x = 4 thỏa mãn ĐKXĐ thì

\(A=\frac{-3}{4-3}=-3\)

d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)

\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)

Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)

A= \(\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)

\(=\dfrac{-6}{\left(x+2\right)\left(x-2\right)}:\dfrac{6}{x+2}\)

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

b) \(\left|x\right|=\dfrac{1}{2}\)

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

Thay \(x=\dfrac{1}{2}\) (Thỏa mãn ĐKXĐ \(x\ne2;x\ne-2\) )

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

Thay \(x=\dfrac{-1}{2}\) (Thỏa mãn ĐKXĐ \(x\ne2;x\ne-2\) )

\(A=\dfrac{-1}{\dfrac{-1}{2}-2}=\dfrac{2}{5}\)

c) \(\dfrac{-1}{x-2}< 0\)

\(\Rightarrow x-2>0\)

\(\Leftrightarrow x>2\)

Vậy x > 2 để A < 0

Câu 1: 

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

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

b: Để A=x/6 thì \(\dfrac{x-1}{x+1}=\dfrac{x}{6}\)

\(\Leftrightarrow x^2+x-6x+6=0\)

=>x=3 hoặc x=2

A=(xx2−4+22−x+1x+2):((x−2)+10−x2x+2)A=(xx2−4+22−x+1x+2):((x−2)+10−x2x+2)

=(x(x−2)(x+2)−2x−2+1x+2):(x−2)(x+2)+10−x2x+2(x(x−2)(x+2)−2x−2+1x+2):(x−2)(x+2)+10−x2x+2

=x−2(x+2)+x−2(x−2)(x+2):x2−4+10−x2x+2x−2(x+2)+x−2(x−2)(x+2):x2−4+10−x2x+2

=x−2(x+2)+x−2(x−2)(x+2)−6(x−2)(x+2):6x+2:x2−4+10−x2x+2x−2(x+2)+x−2(x−2)(x+2)−6(x−2)(x+2):6x+2:x2−4+10−x2x+2

=−6(x−2)(x+2).x+26−6(x−2)(x+2).x+26

=−1x−2=12−x−1x−2=12−x

b) Giá trị của A tại |x|=12|x|=12

Nếu x=12x=12 thì A=12−12=132=23A=12−12=132=23

Nếu x=−12x=−12 thì A=12−(−12)=12+12=152=25A=12−(−12)=12+12=152=25

c) A < 0 khi 2 – x < 0 hay x > 2

d)Để A=0

Dễ thế bạn , bạn quy đồng lên lấy (x-2)(x+2) chung rồi bạn tính ra là đc

B3;a,ĐKXĐ:\(x\ne\pm4\)

A=\(\left(\dfrac{4}{x-4}-\dfrac{4}{x+4}\right)\dfrac{x^2+8x+16}{32}=\left(\dfrac{4x+16}{x^2-16}-\dfrac{4x-16}{x^2-16}\right)\dfrac{x^2+2.4x+4^2}{32}=\left(\dfrac{4x+16-4x+16}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\left(\dfrac{32}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\dfrac{32\left(x+4\right)^2}{32.\left(x-4\right)\left(x+4\right)}=\dfrac{x+4}{x-4}\\ \\ \\ \\ \\ \\ b,Tacó\dfrac{x+4}{x-4}=\dfrac{1}{3}\Leftrightarrow3x+12=x-4\Leftrightarrow x=-8\left(TM\right)c,TAcó\dfrac{x+4}{x-4}=3\Leftrightarrow x+4=3x-12\Leftrightarrow x=8\left(TM\right)\)

\(\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)

= \(\left[-\dfrac{1}{x-1}+\dfrac{2}{x+1}+\dfrac{5-x}{\left(x+1\right)\left(x-1\right)}\right]:\dfrac{1-2x}{x^2-1}\)

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

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

Để A nhận giá trị nguyên

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

Để \(\left|A\right|=A\)

=> 1 - 2x > 0

<=> -2x > -1

<=> x < 1/2

a: ĐKXĐ: x<>3; x<>-3; \(x\ne-5\pm\sqrt{34}\)

b: \(=\dfrac{x^2+5x+6+5x-15}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{2x\left(x-3\right)\left(x+3\right)}{x^2+10x-9}\)

=2x

c: Khi x=1/2 thì A=2*1/2=1