\(a,\left(1\right)=\dfrac{x}{2x^2-3x+10x-15}=\dfrac{x}{\left(2x-3\right)\left(x+5\right)}=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(2x-3\right)\left(x+5\right)}\\ \left(2\right)=\dfrac{x+2}{x^2-2x+5x-10}=\dfrac{x+2}{\left(x-2\right)\left(x+5\right)}=\dfrac{\left(x+2\right)\left(2x-3\right)}{\left(x-2\right)\left(2x-3\right)\left(x+5\right)}\\ \left(3\right)=\dfrac{1}{x+5}=\dfrac{\left(x-2\right)\left(2x-3\right)}{\left(x-2\right)\left(2x-3\right)\left(x+5\right)}\)
\(b,\left(1\right)=\dfrac{-1}{\left(x-2\right)\left(x-1\right)}=\dfrac{\left(x-3\right)\left(x+6\right)}{\left(x-1\right)\left(x-3\right)\left(x+6\right)\left(x-2\right)}\\ \left(2\right)=\dfrac{1}{\left(x+6\right)\left(x-1\right)}=\dfrac{\left(x-3\right)\left(x-2\right)}{\left(x-1\right)\left(x-3\right)\left(x+6\right)\left(x-2\right)}\\ \left(3\right)=\dfrac{-1}{\left(x-1\right)\left(x-3\right)}=\dfrac{-\left(x-2\right)\left(x+6\right)}{\left(x-1\right)\left(x-3\right)\left(x+6\right)\left(x-2\right)}\)
\(c,\left(1\right)=\dfrac{3}{\left(x-1\right)\left(x^2+x+1\right)}\\ \left(2\right)=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\\ \left(3\right)=\dfrac{x\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\\ d,\left(1\right)=\dfrac{x}{\left(x-y-z\right)\left(x-y+z\right)}=\dfrac{x\left(x+y-z\right)}{\left(x-y-z\right)\left(x+y-z\right)\left(x-y+z\right)}\\ \left(2\right)=\dfrac{y}{\left(x-y+z\right)\left(x+y-z\right)}=\dfrac{y\left(x-y-z\right)}{\left(x-y-z\right)\left(x+y-z\right)\left(x-y+z\right)}\\ \left(3\right)=\dfrac{z}{\left(x-y-z\right)\left(x+y-z\right)}=\dfrac{z\left(x-y+z\right)}{\left(x-y-z\right)\left(x+y-z\right)\left(x-y+z\right)}\)
