1: ĐKXĐ: x∉{0;-1}

Ta có: \(\frac{x-1}{x}+\frac{1-2x}{x\left(x+1\right)}=\frac{1}{x+1}\)

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

=>\(\left(x-1\right)\left(x+1\right)+1-2x=x\)

=>\(x^2-1+1-2x-x=0\)

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

=>x(x-3)=0

=>\(\left[\begin{array}{l}x=0\left(loại\right)\\ x=3\left(nhận\right)\end{array}\right.\)

2: ĐKXĐ: x∉{0;4}

ta có: \(\frac{5}{x}+\frac{x-3}{x-4}=\frac{x^2-10}{x\left(x-4\right)}\)

=>\(\frac{5\left(x-4\right)+x\left(x-3\right)}{x\left(x-4\right)}=\frac{x^2-10}{x\left(x-4\right)}\)

=>\(5\left(x-4\right)+x\left(x-3\right)=x^2-10\)

=>\(5x-20+x^2-3x=x^2-10\)

=>2x-20=-10

=>2x=10

=>x=5(nhận)

3: ĐKXĐ: x∉{0;3}

Ta có: \(\frac{x+3}{x-3}=\frac{3}{x^2-3x}+\frac{1}{x}\)

=>\(\frac{x+3}{x-3}=\frac{3}{x\left(x-3\right)}+\frac{1}{x}\)

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

=>\(x\left(x+3\right)=3+x-3=x\)

=>\(x^2+3x-x=0\)

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

=>x(x+2)=0

=>\(\left[\begin{array}{l}x=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(loại\right)\\ x=-2\left(nhận\right)\end{array}\right.\)

4: ĐKXĐ: x∉{0;3}

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

=>\(\frac{3}{x\left(x-3\right)}+\frac{1}{x}=\frac{x+4}{x-3}\)

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

=>\(x=x\left(x+4\right)\)

=>x(x+4)-x=0

=>x(x+3)=0

=>\(\left[\begin{array}{l}x=0\left(loại\right)\\ x=-3\left(nhận\right)\end{array}\right.\)

5: ĐKXĐ: x∉{0;4}

ta có: \(\frac{x+4}{x-4}-\frac{1}{x}=\frac{4}{x^2-4x}\)

=>\(\frac{x+4}{x-4}-\frac{1}{x}=\frac{4}{x\left(x-4\right)}\)

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

=>\(x\left(x+4\right)-x+4=4\)

=>\(x^2+4x-x=0\)

=>\(x^2+3x=0\)

=>x(x+3)=0

=>\(\left[\begin{array}{l}x=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(loại\right)\\ x=-3\left(nhận\right)\end{array}\right.\)

6: ĐKXĐ: x∉{3;-1}

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

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

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

=>\(x^2+x+x^2-3x=2x^2-4\)

=>-2x=-4

=>x=2(nhận)

7: ĐKXĐ: x∉{0;2}

ta có: \(\frac{x+2}{x-2}-\frac{6}{x}=\frac{9}{x^2-2x}\)

=>\(\frac{x+2}{x-2}-\frac{6}{x}=\frac{9}{x\left(x-2\right)}\)

=>\(\frac{x\left(x+2\right)-6\left(x-2\right)}{x\left(x-2\right)}=\frac{9}{x\left(x-2\right)}\)

=>x(x+2)-6(x-2)=9

=>\(x^2+2x-6x+12-9=0\)

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

=>(x-1)(x-3)=0

=>\(\left[\begin{array}{l}x-1=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\left(nhận\right)\\ x=3\left(nhận\right)\end{array}\right.\)

8: ĐKXĐ: x∉{0;2}

ta có: \(\frac{2}{x^2-2x}+\frac{1}{x}=\frac{x+2}{x-2}\)

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

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

=>x(x+2)=x

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

=>x(x+2-1)=0

=>x(x+1)=0

=>\(\left[\begin{array}{l}x=0\left(loại\right)\\ x=-1\left(nhận\right)\end{array}\right.\)