\(B=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{x+2}\right):\frac{x^2-3x}{2x^2-x^3}\left(ĐKXĐ:x\ne2;-2;0\right)\)

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

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

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

\(B=\frac{-5x^2-7x}{\left(x+2\right)}.\frac{-x}{x-3}\)

\(B=\frac{\left(-5x^2-7x\right)-x}{\left(x+2\right)\left(x-3\right)}\)

\(B=\frac{5x^3+7x^2}{\left(x+2\right)\left(x+3\right)}\)

\(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)   ĐKXD: \(x\ne\pm2,x\ne0,x\ne3\)

\(\Leftrightarrow\left(\frac{2+x}{2-x}+\frac{4x^2}{\left(2-x\right)\left(2+x\right)}-\frac{2-x}{2+x}\right):\left(\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\right)\)

\(\Leftrightarrow\left(\frac{4+4x+x^2+4x^2-4+4x-x^2}{\left(2-x\right)\left(2+x\right)}\right):\left(\frac{x-3}{x\left(2-x\right)}\right)\)

\(\Leftrightarrow\left(\frac{4x^2+8x}{\left(2-x\right)\left(2+x\right)}\right)\cdot\left(\frac{x\left(2-x\right)}{x-3}\right)\)

\(\Leftrightarrow\frac{4x\left(x+2\right)}{\left(2-x\right)\left(2+x\right)}\cdot\frac{x\left(2-x\right)}{x-3}\)

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

b, Để A>0 thì \(\frac{4x^2}{x-3}>0\)

\(\Rightarrow4x^2>0\)

\(\Rightarrow x>0\)

c, Ta có

\(\left|x-7\right|=4\)

\(\Rightarrow\orbr{\begin{cases}x-7=4\\x-7=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=3\left(l\right)\end{cases}}}\)

 Với \(x=11\Rightarrow\frac{4\cdot11^2}{11-3}=\frac{121}{2}\)

\(DKXD:x\ne\pm2;x\ne3;x\ne\frac{3}{2};x\ne0\)

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

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

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

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

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

b

Xét hơi bị nhiều TH nhá:(

Để \(A>0\) thì \(\frac{4x\left(2x-3\right)}{\left(2+x\right)\left(x-3\right)}>0\)

TH1:\(4x\left(2x-3\right)>0;\left(2+x\right)\left(x-3\right)>0\)

\(TH2:4x\left(2x-3\right)< 0;\left(2+x\right)\left(x-3\right)< 0\)

Bạn tự xét nốt nhá!

c

\(\left|x-7\right|=4\Rightarrow x-7=4;x-7=-4\)

\(\Rightarrow x=11;x=3\)

Thay vào .....

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

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

\(=\dfrac{x^2+5x+6+x^2+4x+3+x^2+3x+2}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)

\(=\dfrac{3x^2+12x+11}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)

a, ĐKXĐ: \(x\ne\pm3\)

\(A=\frac{x\left(x-3\right)+2x\left(x+3\right)-3x^2-12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)

\(=\frac{3x-12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}=\frac{3x-12}{3x+9}\)

b, \(x=-4\Rightarrow A=\frac{3.\left(-4\right)-12}{3.\left(-4\right)+9}=8\)

c, \(A\in Z\Rightarrow3x-12⋮\left(3x+9\right)\Rightarrow3x+9-21⋮\left(3x+9\right)\Rightarrow21⋮\left(3x+9\right)\)

\(\Rightarrow3x+9\inƯ\left(21\right)=\left\{\pm1;\pm3;\pm7;\pm21\right\}\)

Mà \(3x+9⋮3\Rightarrow3x+9\in\left\{-21;-3;3;21\right\}\Rightarrow x\in\left\{-10;-4;-2;4\right\}\) (thỏa mãn điều kiện)

a, ĐỂ A xác định : 

\(\Rightarrow\hept{\begin{cases}x+3\ne0\\x-3\ne0\\x^2-9\ne0\end{cases}}\Rightarrow x\ne\pm3.\)

\(A=\left(\frac{x}{x+3}+\frac{2x}{x-3}-\frac{3x^2+12}{\left(x+3\right)\left(x-3\right)}\right):\frac{3}{x-3}\)

\(A=\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3x^2+12}{\left(x-3\right)\left(x+3\right)}:\frac{3}{x-3}\)

\(A=\frac{x^2-3x+2x^2+6x-3x^2+12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)

\(A=\frac{3x+12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)

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

b

a) \(A=\left(\frac{x}{x+3}+\frac{2x}{x-3}-\frac{3x^2+12}{x^2-9}\right):\frac{3}{x-3}\)

\(A=\left[\frac{x}{x+3}+\frac{2x}{x-3}-\frac{3x^2+12}{\left(x-3\right)\left(x+3\right)}\right]:\frac{3}{x-3}\)

A xác định \(\Leftrightarrow\hept{\begin{cases}x+3\ne0\\x-3\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne-3\\x\ne3\end{cases}}}\)

b) \(A=\left[\frac{x}{x+3}+\frac{2x}{x-3}-\frac{3x^2+12}{\left(x-3\right)\left(x+3\right)}\right]:\frac{3}{x-3}\)

\(A=\left[\frac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3x^2+12}{\left(x-3\right)\left(x+3\right)}\right]:\frac{3}{x-3}\)

\(A=\left[\frac{x^2-3x+2x^2+6x-3x^2-12}{\left(x+3\right)\left(x-3\right)}+\right]:\frac{3}{x-3}\)

\(A=\left[\frac{3x-12}{\left(x+3\right)\left(x-3\right)}\right].\frac{x-3}{3}\)

\(A=\left[\frac{3\left(x-4\right)}{\left(x+3\right)\left(x-3\right)}\right].\frac{x-3}{3}\)

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

Với \(x=-4\)

\(\Rightarrow A=\frac{-4-4}{-4+3}=-\frac{8}{-1}=8\)

Vậy \(A=8\)tại \(x=-4\)

c) \(A=\frac{x-4}{x+3}=\frac{x+3-7}{x+3}=1-\frac{7}{x+3}\)

Có \(1\in Z\)

Để \(A\in Z\Rightarrow\frac{7}{x+3}\in Z\)

Có: \(x\in Z\Rightarrow x+3\in Z\Rightarrow\frac{7}{x+3}\in Z\Leftrightarrow\left(x+3\right)\in\text{Ư}\left(7\right)=\left\{\pm1;\pm7\right\}\)

b tự lập bảng nhé~

a, ĐKXĐ : \(\hept{\begin{cases}2-x\ne0\\x^2-4\ne0\\2+x\ne0\end{cases}}\)hoặc \(2x^2-x^3\ne0\)hay \(x\ne\pm2;0\)

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

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

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

\(=\frac{-4x^2-6x+3}{\left(x-2\right)\left(x+2\right)}.\frac{-x\left(x-2\right)}{x-3}=\frac{\left(-4x^2-6x+3\right)\left(-x\right)}{\left(x+2\right)\left(x-3\right)}=\frac{4x^3+6x^2-3x}{\left(x+2\right)\left(x-3\right)}\)

b, Ta có : A > 0 hay \(\frac{4x^3+6x^2-3x}{\left(x+2\right)\left(x-3\right)}>0\)

\(\Leftrightarrow x\left(4x^2+6x-3\right)>0\)

\(\Leftrightarrow4x^2+6x-3>0\) bạn xem lại bài mình có chỗ nào sai ko nhé !!! 

c, Ta có : \(\left|x-7\right|=4\Rightarrow\orbr{\begin{cases}x-7=4\\x-7=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=3\end{cases}}}\)

TH1 : Thay x = 11 vào phân thức trên : ... 

TH2 : Thay x = 3 vào phân thức trên : .... tự làm 

Bn ơi s bài này bn ko đổi dấu cái ở giữa nhỉ ns dễ hơn