\(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)   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 \(x=11\Rightarrow\frac{4\cdot11^2}{11-3}=\frac{121}{2}\)

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

\(ĐK:x\ne\pm1;x\ne0;x\ne3\)

Với \(x\ne\pm1;x\ne0;x\ne3\)thì\(M=\frac{x^3+2x^2-x-2}{x^3-2x^2-3x}\left[\frac{\left(x+2\right)^2-x^2}{4x^2-4}-\frac{3}{x^2-x}\right]=\frac{x^2\left(x+2\right)-\left(x+2\right)}{\left(x^3-x\right)-\left(2x^2+2x\right)}\left[\frac{x^2+4x+4-x^2}{4x^2-4}-\frac{3}{x\left(x-1\right)}\right]\)\(=\frac{\left(x-1\right)\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)\left(x-1\right)-2x\left(x+1\right)}\left[\frac{4\left(x+1\right)}{4\left(x+1\right)\left(x-1\right)}-\frac{3}{x\left(x-1\right)}\right]=\frac{\left(x-1\right)\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x^2-3x\right)}\left[\frac{1}{x-1}-\frac{3}{x\left(x-1\right)}\right]\)\(=\frac{\left(x-1\right)\left(x+2\right)}{x\left(x-3\right)}.\frac{x-3}{x\left(x-1\right)}=\frac{x+2}{x^2}\)

M = 3 \(\Leftrightarrow\frac{x+2}{x^2}=3\Leftrightarrow3x^2-x-2=0\Leftrightarrow\left(x-1\right)\left(3x+2\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{-2}{3}\end{cases}}\)

Mà \(x\ne1\)(theo điều kiện) nên x =^-2/₃