\(a.\) Ta có: 

 \(MTC:\)  \(\left(x+1\right)\left(x+2\right)\)

 Do đó

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

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

\(b.\)  Ta có: 

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

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

nên  \(MTC:\)  \(x\left(x-1\right)\left(x+1\right)\)

Do đó:

\(\frac{5}{x^2+x}=\frac{5}{x\left(x+1\right)}=\frac{5\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)

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

\(c.\)  Ta có:

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

\(2x^2-8x=2x\left(x-4\right)\)

nên  \(MTC:\)  \(2x\left(x-1\right)\left(x-4\right)\)

Do đó: 

\(\frac{4}{x^2-5x+4}=\frac{4}{\left(x-1\right)\left(x-4\right)}=\frac{8x}{2x\left(x-1\right)\left(x-4\right)}\)

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

Làm nốt d :P

\(\frac{x+3}{2x^2-15x-8};\frac{3}{x^2-8x}\)

Ta có : \(2x^2-15x-8=\left(2x+1\right)\left(x-8\right)\)

\(x^2-8x=x\left(x-8\right)\)

MTC : \(x\left(x-8\right)\left(2x+1\right)\)

\(\frac{x+3}{2x^2-15x-8}=\frac{x+3}{\left(2x+1\right)\left(x-8\right)}=\frac{x^2+3x}{x\left(x-8\right)\left(2x+1\right)}\)

\(\frac{3}{x^2-8x}=\frac{3}{x\left(x-8\right)}=\frac{6x+3}{x\left(x-8\right)\left(2x+1\right)}\)

Khi thay  \(xyz=1\) vào biểu thức  \(\frac{1}{1+x+xy}\) thì được \(\frac{1}{1+x+xy}\) (tự chứng minh)

Khi thay  \(xyz=1\) vào biểu thức  \(\frac{1}{1+y+yz}\) thì được:

\(\frac{xyz}{xyz+y+yz}=\frac{xyz}{y\left(xz+z+1\right)}=\frac{xz}{xz+z+1}=\frac{xz}{xz+z+xyz}=\frac{xz}{z\left(1+x+xy\right)}=\frac{x}{1+x+xy}\)

Khi thay  \(xyz=1\) vào biểu thức  \(\frac{1}{1+z+zx}\) thì được:

\(\frac{xyz}{xyz+z+xz}=\frac{xyz}{z\left(1+x+xy\right)}=\frac{xy}{1+x+xy}\)

Do đó:  \(\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}=\frac{1}{1+x+xy}+\frac{x}{1+x+xy}+\frac{xy}{1+x+xy}=\frac{1+x+xy}{1+x+xy}=1\)

câu a nè

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

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

câu c nè

\(\frac{x^2-3x+5}{x+1}=\frac{\left(x^2+2x+1\right)-5x+4}{x+1}=\frac{\left(x+1\right)^2-5\left(x+1\right)+9}{x+1}\)

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