\(a,=\dfrac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)};\dfrac{5-3x}{x^2-9}=\dfrac{2x\left(5-3x\right)}{2x\left(x-3\right)\left(x+3\right)}\\ b,\dfrac{x+1}{x-x^2}=\dfrac{2\left(x+1\right)\left(1-x\right)}{2x\left(1-x\right)^2};\dfrac{x+2}{2-4x+2x^2}=\dfrac{x\left(x+2\right)}{2x\left(1-x\right)^2}\\ c,\dfrac{4x^2-3x+5}{x^3-1}=\dfrac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\\ \dfrac{2x}{x^2+x+1}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)};\dfrac{6}{x-1}=\dfrac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\dfrac{7}{5x}=\dfrac{14\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}\\ \dfrac{4}{x-2y}=\dfrac{40x\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}\\ \dfrac{x-y}{8y^2-2x^2}=\dfrac{5x\left(y-x\right)}{10x\left(x-2y\right)\left(x+2y\right)}\\ e,\dfrac{5x^3}{x^3+6x^2+12x+8}=\dfrac{10x^3}{2\left(x+2\right)^3}\\ \dfrac{4x}{x^2+4x+4}=\dfrac{8x\left(x+2\right)}{2\left(x+2\right)^3};\dfrac{3}{2x+4}=\dfrac{3\left(x+2\right)^2}{2\left(x+2\right)^3}\)
\(f,\dfrac{x}{x^3-1}=\dfrac{x^2}{x\left(x-1\right)\left(x^2+x+1\right)};\dfrac{x+1}{x^2-x}=\dfrac{\left(x+1\right)\left(x^2+x+1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}\\\dfrac{x-1}{x^2+x+1}=\dfrac{x\left(x-1\right)^2}{x\left(x-1\right)\left(x^2+x+1\right)}\\ i,PT\left(1\right)=\dfrac{x}{\left(x-y-z\right)\left(x-y+z\right)}=\dfrac{x\left(x+y-z\right)}{\left(x-y-z\right)\left(x-y+z\right)\left(x+y-z\right)}\\ PT\left(2\right)=\dfrac{y}{\left(x-y+z\right)\left(x+y-z\right)}=\dfrac{y\left(x-y-z\right)}{\left(x-y-z\right)\left(x-y+z\right)\left(x+y-z\right)}\\ PT\left(3\right)=\dfrac{z}{\left(x-y-z\right)\left(x+y-z\right)}=\dfrac{z\left(x-y+z\right)}{\left(x-y-z\right)\left(x+y-z\right)\left(x-y+z\right)}\)
