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

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

\(\frac{x+2}{4x-x^2-3}=\frac{-\left(x+2\right)}{x^2-4x+3}=\frac{\left(-x-2\right)\left(2x+5\right)}{\left(x-1\right)\left(x-3\right)\left(2x+5\right)}=\frac{-2x^2-9x-10}{\left(x-1\right)\left(x-3\right)\left(2x+5\right)}\)

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

Bài 1:

a: \(\frac{1}{2x^3y}=\frac{1\cdot6\cdot yz^3}{2x^3y\cdot6yz^3}=\frac{6yz^3}{12x^3y^2z^3}\)

\(\frac{2}{3xy^2z^3}=\frac{2\cdot4\cdot x^2}{3xy^2z^3\cdot4x^2}=\frac{8x^2}{12x^3y^2z^3}\)

\(\frac{5}{4yz}=\frac{5\cdot3\cdot x^3\cdot y\cdot z^2}{4yz\cdot3x^3yz^2}=\frac{15x^3yz^2}{12x^3y^2z^3}\)

b: \(\frac{x+1}{10x^3-40x}=\frac{x+1}{10x\left(x^2-4\right)}=\frac{x+1}{10x\left(x-2\right)\left(x+2\right)}\)

\(=\frac{\left(x+1\right)\cdot4\cdot x}{4x\cdot10x\cdot\left(x+2\right)\left(x-2\right)}=\frac{4x^2+4x}{40x^2\left(x+2\right)\left(x-2\right)}\)

\(\frac{5}{8x^3+16x^2}=\frac{5x}{8x^2\left(x+2\right)}\)

\(=\frac{5x\cdot5\cdot\left(x-2\right)}{8x^2\left(x+2\right)\cdot5\cdot\left(x-2\right)}=\frac{25x^2-50x}{40x^2\left(x+2\right)\left(x-2\right)}\)

Bài 2:

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

\(=\frac{\left(x-2\right)\cdot4x\cdot\left(x^2+x+1\right)}{3x\left(x-1\right)\cdot4x\cdot\left(x^2+x+1\right)}=\frac{\left(4x^2-8x\right)\left(x_{}^2+x+1\right)}{12x^2\left(x-1\right)\left(x^2+x+1\right)}\)

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

\(=\frac{\left(x^2-2\right)\cdot3}{4x^2\left(x-1\right)\left(x^2+x+1\right)\cdot3}=\frac{3x^2-6}{12x^2\left(x-1\right)\left(x^2+x+1\right)}\)

Bài 2:

a: \(\dfrac{1}{2x^3y}=\dfrac{6yz^3}{12x^3y^2z^3}\)

\(\dfrac{2}{3xy^2z^3}=\dfrac{2\cdot4x^2}{12x^3y^2z^3}=\dfrac{8x^2}{12x^3y^2z^3}\)

a) MTC: \(12x^3y^3\)

\(\dfrac{3}{4x^3y^2}=\dfrac{3\cdot3y}{4x^3y^2\cdot3y}=\dfrac{9y}{12x^3y^3}\)

\(\dfrac{2}{3xy^3}=\dfrac{2\cdot4x^2}{3xy^3\cdot4x^2}=\dfrac{8x^2}{12x^3y^3}\)

b) MTC: \(x\left(x-3\right)^2\)

\(\dfrac{5}{x^2-6x+9}=\dfrac{5}{\left(x-3\right)^2}=\dfrac{5x}{x\left(x-3\right)^2}\)

\(\dfrac{3}{x^2-3x}=\dfrac{3}{x\left(x-3\right)}=\dfrac{3\left(x-3\right)}{x\left(x-3\right)^2}=\dfrac{3x-9}{x\left(x-3\right)^2}\)

Ta có \(\frac{2}{x^3-y^3}=\frac{2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

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

\(\frac{1}{x+y}\)  giữ nguyên

MTC: \(\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)

Các nhân tử phụ tương ứng là : \(\left(x+y\right);\left(x-y\right)\left(x^2+xy+y^2\right);\left(x^2+xy+y^2\right)\)

Ta có:

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

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

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

cho mình hỏi là giữa khác phân số với nhua là phải có dấu như là công, trừ, nhân hay chia chứ? 

\(\frac{-3}{x^2+6x+8}=\frac{-3}{x\left(x+2\right)+4\left(x+2\right)}=\frac{-3}{\left(x+2\right)\left(x+4\right)}=\frac{-3x+12}{\left(x+2\right)\left(x+4\right)\left(x-4\right)}\)

\(\frac{5}{x^2-16}=\frac{5}{\left(x-4\right)\left(x+4\right)}=\frac{5x+10}{\left(x+2\right)\left(x-4\right)\left(x+4\right)}\)

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