\(a,\left(1\right)=\dfrac{x-1}{2\left(x+1\right)}=\dfrac{\left(x-1\right)^2}{2\left(x+1\right)\left(x-1\right)}\\ \left(2\right)=\dfrac{x+1}{2\left(x-1\right)}=\dfrac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}\\ \left(3\right)=\dfrac{-1}{\left(x-1\right)\left(x+1\right)}=\dfrac{-2}{2\left(x-1\right)\left(x+1\right)}\\ b,\left(1\right)=\dfrac{\left(2x-1\right)\left(x^2-ax+a^2\right)}{\left(x+a\right)\left(x^2-ax+a^2\right)}\\ \left(2\right)=\dfrac{x-a}{x^2-ax+a^2}=\dfrac{\left(x-a\right)\left(x+a\right)}{\left(x+a\right)\left(x^2-ax+a^2\right)}\\ \left(3\right)=\dfrac{2x^2-1}{\left(x+a\right)\left(x^2-ax+a^2\right)}\)
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