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a) đK: \(x\ne0;2\)

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

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

b) Thay x = -2 (TMDK) vào B, ta có:

\(B=\dfrac{4-3.\left(-2\right)}{4}=\dfrac{4+6}{4}=\dfrac{5}{2}\)

c) Để \(\left|B\right|-2x=5\)

<=> \(\left|\dfrac{4-3x}{4}\right|-2x=5\)

TH1: \(x\le\dfrac{4}{3}\)

<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{4-3x}{4}\)

PT <=> \(\dfrac{4-3x}{4}-2x=5\)

<=> \(\dfrac{4-3x-8x}{4}=5\)

<=> \(4-11x=20\)

<=> x = \(\dfrac{-16}{11}\) (Tm)

TH2: \(x>\dfrac{4}{3}\)

<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{3x-4}{4}\)

PT <=> \(\dfrac{3x-4}{4}-2x=5\)

<=> \(\dfrac{3x-4-8x}{4}=5\)

<=> \(-5x-4=20\)

<=> \(x=\dfrac{-24}{5}\left(l\right)\)

d) Xét (2-x)B = \(\dfrac{\left(2-x\right)\left(4-3x\right)}{4}\)  = \(\dfrac{3x^2-10x+8}{4}\)

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

Mà \(3\left(x-\dfrac{5}{3}\right)^2\ge\) 0

=> (2-x)B \(\ge\dfrac{\dfrac{-1}{3}}{4}=\dfrac{-1}{12}\)

Dấu "=" <=> x = \(\dfrac{5}{3}\left(tm\right)\)

e) Số nguyên âm lớn nhất là -1

Để B = -1

<=> \(\dfrac{4-3x}{4}=-1\)

<=> 4 - 3x = -4
<=> \(x=\dfrac{8}{3}\left(tm\right)\)

g) 

TH1: \(x\le\dfrac{4}{3}\)

<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{4-3x}{4}\)

BDT <=> \(\dfrac{4-3x}{4}< 2x-4\)

<=> \(4-3x< 8x-16\)

<=> \(x>\dfrac{20}{11}\left(l\right)\)

TH2: \(x>\dfrac{4}{3}\)

<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{3x-4}{4}\)

BDT <=> \(\dfrac{3x-4}{4}< 2x-4\)

<=> \(3x-4< 8x-16\)

<=> x > \(\dfrac{12}{5}\)

KHDK: \(x>\dfrac{12}{5}\)

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

\(< =>B=\left(\frac{x-4}{x\left(x-2\right)}+\frac{2x}{x\left(x-2\right)}\right):\left(\frac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)}+\frac{x.x}{x\left(x-2\right)}\right)\)

\(< =>B=\left(\frac{x-4+2x}{x\left(x-2\right)}\right):\left(\frac{x^2-4}{x\left(x-2\right)}+\frac{x^2}{x\left(x-2\right)}\right)\)

\(< =>B=\frac{3x-4}{x\left(x-2\right)}:\frac{x^2-4+x^2}{x\left(x-2\right)}\)

\(< =>B=\frac{3x-4}{x\left(x-2\right)}.\frac{x\left(x-2\right)}{2x^2-4}\)

\(< =>B=\frac{3x-4}{2x^2-4}\)

\(b,\)Với \(x=-2\)thì

 \(B=\frac{3\left(-2\right)-4}{2\left(-2\right)^2-4}=\frac{-6-4}{8-4}=-\frac{10}{4}=-\frac{5}{2}\)

\(ĐKXĐ:x\ne2;x\ne0\)

a

\(B=\left[\frac{x-4}{x\left(x-2\right)}+\frac{2}{x-2}\right]:\left(\frac{x+2}{x}-\frac{x}{x-2}\right)\)

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

\(=\frac{3x-4}{x^2-4-x^2}=-\frac{3x-4}{4}\)

b

\(B=-\frac{3x-4}{4}=-\frac{3\cdot\left(-2\right)-4}{4}=\frac{5}{2}\)

c

\(\left|B\right|-2x=5\Leftrightarrow\left|B\right|=5+2x\)

\(B=-\frac{3x-4}{4}\Leftrightarrow-\frac{3x-4}{4}\ge0\Leftrightarrow x\le\frac{4}{3}\)

\(B=\frac{3x-4}{4}\Leftrightarrow x>\frac{4}{3}\)

Xét các trường hợp của x thì ra nghiệm bạn nhé

d

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

Để ( 2 - x ).B đạt giá trị nhỏ nhất thì ( 2 - x ) ( 3x - 4 ) đạt giá trị lớn nhất

Casio sẽ giúp chúng ta phần này

e

Để B là số nguyên âm lớn nhất hay \(B=-1\Leftrightarrow-\frac{3x-4}{4}=-1\Leftrightarrow x=\frac{8}{3}\)

g

\(\left|B\right|+3< 2x-1\)

Làm hệt như câu c nhé :D 

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

ĐKXĐ : \(x\ne0,x\ne2\)

a) \(B=\left(\frac{x-4}{x\left(x-2\right)}+\frac{2x}{x\left(x-2\right)}\right):\left(\frac{\left(x+2\right)\left(x-2\right)}{x\left(x-2\right)}-\frac{x\cdot x}{x\left(x-2\right)}\right)\)

\(B=\left(\frac{x-4+2x}{x\left(x-2\right)}\right):\left(\frac{x^2-4-x^2}{x\left(x-2\right)}\right)\)

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

\(B=\frac{3x-4}{-4}=\frac{-3x+4}{4}\)

b) Thế x = -2 ( tmđk ) vào B ta được :

\(B=\frac{-3\cdot\left(-2\right)+4}{4}=\frac{10}{4}=\frac{5}{2}\)

c) \(\left|B\right|-2x=5\)

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

\(\Leftrightarrow\frac{-3x+4}{4}-2x=5\)

\(\Leftrightarrow\frac{-3x+4}{4}-\frac{8x}{4}=5\)

\(\Leftrightarrow\frac{-3x+4-8x}{4}=5\)

\(\Leftrightarrow\frac{-11x+4}{4}=5\)

\(\Leftrightarrow-11x+4=20\)

\(\Leftrightarrow-11x=16\)

\(\Leftrightarrow x=-\frac{16}{11}\)

Nhờ các bạn khác làm nốt ạ -.-

Bài 1 : 

a, \(A=\frac{2x^2-4x+8}{x^3+8}=\frac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{2}{x+2}\)

b, Ta có : \(\left|x\right|=2\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

TH1 : Thay x = 2 vào biểu thức trên ta được : 

\(\frac{2}{2+2}=\frac{2}{4}=\frac{1}{2}\)

TH2 : Thay x = -2 vào biểu thức trên ta được : 

\(\frac{2}{-2+2}=\frac{2}{0}\)vô lí 

c, ta có A = 2 hay \(\frac{2}{x+2}=2\)ĐK : \(x\ne-2\)

\(\Rightarrow2x+4=2\Leftrightarrow2x=-2\Leftrightarrow x=-1\)

Vậy với x = -1 thì A = 2 

d, Ta có A < 0 hay \(\frac{2}{x+2}< 0\)

\(\Rightarrow x+2< 0\)do 2 > 0 

\(\Leftrightarrow x< -2\)

Vậy với A < 0 thì x < -2 

e, Để A nhận giá trị nguyên khi \(x+2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

2.

ĐKXĐ : \(x\ne\pm2\)

a. \(B=\frac{x^2-4x+4}{x^2-4}=\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{x-2}{x+2}\)

b. | x - 1 | = 2 <=>\(\hept{\begin{cases}x-1=2\\x-1=-2\end{cases}}\)<=>\(\hept{\begin{cases}x=3\\x=-1\end{cases}}\)

Với x = 3 thì \(B=\frac{3-2}{3+2}=\frac{1}{5}\)

Với x = - 1 thì \(B=\frac{-1-2}{-1+2}=-3\)

Vậy với | x - 1 | = 2 thì B đạt được 2 giá trị là B = 1/5 hoặc B = - 3

c. \(B=\frac{x-2}{x+2}=-1\)<=>\(-\left(x-2\right)=x+2\)

<=> \(-x+2=x+2\)<=>\(-x=x\)<=>\(x=0\)

d. \(B=\frac{x-2}{x+2}< 1\)<=>\(x-2< x+2\)luôn đúng \(\forall\)x\(\ne\pm2\)

e. \(B=\frac{x-2}{x+2}=\frac{x+2-4}{x+2}=1-\frac{4}{x+2}\)

Để B nguyên thì 4/x+2 nguyên => x + 2\(\in\){ - 4 ; - 2 ; - 1 ; 1 ; 2 ; 4 }

=> x \(\in\){ - 6 ; - 4 ; - 3 ; - 1 ; 0 ; 2 }

Đề bài là \(B=\dfrac{\left(x-1\right)^2-4}{\left(2x+1\right)^2-\left(x+2\right)^2}\) hay là \(B=\dfrac{\left(x-1\right)^2-4}{\left(2x+1\right)^2}-\left(x+2\right)^2?\)

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

viết lại biểu thức 

a) \(B=\dfrac{\left(x-1\right)^2-4}{\left(2x+1\right)^2-\left(x+2\right)^2}=\dfrac{\left(x-1-2\right)\left(x-1+2\right)}{\left(2x+1-x-2\right)\left(2x+1+x+2\right)}=\dfrac{\left(x+1\right)\left(x-3\right)}{3\left(x-1\right)\left(x+1\right)}\)  (1)

\(\Rightarrow\) ĐKXĐ: \(x\ne\pm1\)

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

c) Thay \(x=-3;x=1\) vào (2) ta có: \(\left\{{}\begin{matrix}B=\dfrac{-3-3}{3.\left(-3\right)-3}=\dfrac{1}{2}\\B=\dfrac{1-3}{3.1-3}=0\end{matrix}\right.\)

d) \(B=5\Rightarrow\dfrac{x-3}{3x-3}=5\Leftrightarrow x-3=15x-15\Leftrightarrow x=\dfrac{6}{7}\)

a) Ta có: \(P=\dfrac{x-2}{x^2-1}-\dfrac{x+2}{x^2+2x+1}\cdot\dfrac{1-x^2}{2}\)

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

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

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

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

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

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

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

\(=\dfrac{-x^3+5x-6}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{-\left(x^3-5x+6\right)}{2\left(x-1\right)\left(x+1\right)}\)

a: \(A=\left[\left(\dfrac{4x}{x+2}+\dfrac{8x^2}{4-x^2}\right)\right]:\left[\dfrac{x-1}{x^2-2x}-\dfrac{2}{x}\right]\)

\(=\left(\dfrac{4x}{x+2}-\dfrac{8x^2}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{x-1}{x\left(x-2\right)}-\dfrac{2}{x}\right)\)

\(=\dfrac{4x\left(x-2\right)-8x^2}{\left(x+2\right)\left(x-2\right)}:\dfrac{x-1-2\left(x-2\right)}{x\left(x-2\right)}\)

\(=\dfrac{-8x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-2\right)}{x-1-2x+4}\)

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

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

b: \(x^2+2x=15\)

=>\(x^2+2x-15=0\)

=>(x+5)(x-3)=0

=>\(\left[{}\begin{matrix}x+5=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)

Thay x=-5 vào A, ta được:

\(A=\dfrac{8\cdot\left(-5\right)^2}{\left(-5-3\right)\left(-5+2\right)}=\dfrac{8\cdot25}{\left(-8\right)\cdot\left(-3\right)}=\dfrac{25}{3}\)

c: |A|>A

=>A<0

=>\(\dfrac{8x^2}{\left(x-3\right)\left(x+2\right)}< 0\)

=>(x-3)(x+2)<0

TH1: \(\left\{{}\begin{matrix}x-3>0\\x+2< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>3\\x< -2\end{matrix}\right.\)

=>\(x\in\varnothing\)

TH2: \(\left\{{}\begin{matrix}x-3< 0\\x+2>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< 3\\x>-2\end{matrix}\right.\)

=>-2<x<3

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}-2< x< 3\\x\notin\left\{0;2\right\}\end{matrix}\right.\)