a) đkxđ: \(x\ne\pm2\)

Ta có: \(\frac{x+2}{x-2}-\frac{x-2}{x+2}=\frac{4}{x^2-4}\)

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

\(\Rightarrow8x=4\)\(\Rightarrow x=\frac{1}{2}\)

b) đkxđ: \(x\ne\left\{1;-3\right\}\)

PT \(\Leftrightarrow\frac{\left(x+1\right)\left(x+3\right)-\left(x+2\right)\left(x-1\right)+4}{\left(x-1\right)\left(x+3\right)}=0\)

\(\Rightarrow x^2+4x+3-x^2-x+2+4=0\)

\(\Leftrightarrow3x+10=0\Rightarrow x=-\frac{10}{3}\)

c) đkxđ: \(x\ne\left\{0;-1\right\}\)

\(PT\Leftrightarrow\frac{\left(x-1\right)\left(x+1\right)+1-2x}{x\left(x+1\right)}=\frac{x}{x\left(x+1\right)}\)

\(\Rightarrow x^2-1+1-2x=x\)

\(\Leftrightarrow x^2-3x=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=3\left(tm\right)\end{cases}}\)

d) đkxđ: \(x\ne\left\{1;5\right\}\)

\(PT\Leftrightarrow\frac{x-1-3}{\left(x-1\right)\left(x-5\right)}=\frac{5\left(x-5\right)}{\left(x-1\right)\left(x-5\right)}\)

\(\Rightarrow x-4=5x-25\)

\(\Leftrightarrow4x=21\Rightarrow x=\frac{21}{4}\)

e) đkxđ: \(x\ne\left\{0;-2;2\right\}\)

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

\(\Rightarrow-x+2=x^2-6x+8\)

\(\Leftrightarrow x^2-5x+6=0\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

\(\frac{1}{x^2}+\frac{2}{x^2+1}+\frac{3}{x^2+2}+\frac{4}{x^2+3}=4\)

\(\Leftrightarrow\frac{1}{x^2}-1+\frac{2}{x^2+1}-1+\frac{3}{x^2+2}-1+\frac{4}{x^2+3}-1=0\)

\(\Leftrightarrow\frac{1-x^2}{x^2}+\frac{1-x^2}{x^2+1}+\frac{1-x^2}{x^2+2}+\frac{1-x^2}{x^2+3}=0\)

\(\Leftrightarrow1-x^2=0\Leftrightarrow x=\pm1\)

\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{2\left(x+2\right)^2}{x^6-1}\)(ĐK: \(x\ne\pm1\))

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

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

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

\(\Rightarrow x^2-1=\left(x+2\right)^2\)

\(\Leftrightarrow x=-\frac{5}{4}\)(thử lại thỏa mãn).

1siDdG6 vào thống kê nhé 

mình làm mẫu 1 câu nhé !

\(\frac{x+2}{x-2}-\frac{x-2}{x+2}=\frac{4}{x^2-4}\)ĐK : \(x\ne\pm2\)

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

Khử mẫu : \(x^2+2x+1-x^2+2x-1=4\)

\(\Leftrightarrow4x=4\Leftrightarrow x=1\)( tmđk )

Vậy tập nghiệm của phương trình là S = { 1 } 

e, \(\frac{2}{4-x^2}+\frac{1}{x^2-2x}=\frac{x-4}{x^2+2x}\)ĐK : \(x\ne0;\pm2\)

\(\Leftrightarrow\frac{2}{\left(2-x\right)\left(x+2\right)}+\frac{1}{x\left(x-2\right)}-\frac{x-4}{x\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{2x}{x\left(2-x\right)\left(x+2\right)}-\frac{x+2}{x\left(2-x\right)\left(x+2\right)}-\frac{\left(x-4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}=0\)

Khử mẫu : \(2x-x-2-\left(x^2-2x-4x+4\right)=0\)

\(\Leftrightarrow x-2-x^2+6x-4=0\)

\(\Leftrightarrow-x^2+7x-6=0\Leftrightarrow-\left(x-6\right)\left(x-1\right)=0\Leftrightarrow x=6;1\)

Vậy tập nghiệm của phương trình là S = { 1 ; 6 } 

d, \(\frac{1}{x-5}-\frac{3}{x^2-6x+5}=\frac{5}{x-1}\)ĐK : \(x\ne5;1\)

\(\Leftrightarrow\frac{x-1}{\left(x-1\right)\left(x-5\right)}-\frac{3}{\left(x-1\right)\left(x-5\right)}=\frac{5\left(x-5\right)}{\left(x-1\right)\left(x-5\right)}\)

Khử mẫu : \(x-1-3=5x-25\)

\(\Leftrightarrow x-4=5x-25\Leftrightarrow-4x=-21\Leftrightarrow x=\frac{21}{4}\)

f, \(x^3+6x^2+11x+6=0\)

vì : \(Ư\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)( phương pháp nhẩm nghiệm ) 

Ta thấy -1 là nghiệm của phương trình dễ dàng suy ra : x + 1 là nhân tử 

\(\Leftrightarrow\left(x+1\right)\left(x^2+5x+6\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)=0\Leftrightarrow x=-1;-2;-3\)

\(\left(2x-5\right)^3-\left(3x-4\right)^3+\left(x+1\right)^3=0\)

\(\Leftrightarrow8x^3-60x^2+150x-125-\left(27x^3-108x^2+144x-64\right)+x^3+3x^2+3x+1=0\)

\(\Leftrightarrow9x^3-57x^2+153x-124-27x^3+108x^2-144x+64=0\)

\(\Leftrightarrow-18x^3+51x^2+9x-60=0\)

\(\Leftrightarrow\left(3x-4\right)\left(2x-5\right)\left(x+1\right)=0\Leftrightarrow x=\frac{4}{3};\frac{5}{2};-1\)

\(\left(2x-5\right)^3-\left(3x-4\right)^3+\left(x+1\right)^3=0\)

\(\Leftrightarrow8x^3-60x^2+150x-125-27x^3+108x^2-144x+64+x^3+3x^2+3x+1=0\)

\(\Leftrightarrow-18x^3+51x^2+9x-60=0\)

\(\Leftrightarrow-3\left(x+1\right)\left(2x-5\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x+1=0\\2x-5=0\\3x-4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\x=\frac{5}{2}\\x=\frac{4}{3}\end{cases}}}\)