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\(A=\dfrac{\dfrac{x}{x-1}-\dfrac{x+1}{x}}{\dfrac{x}{x+1}-\dfrac{x-1}{x}}=\dfrac{\dfrac{x^2-\left(x^2-1\right)}{x\left(x-1\right)}}{\dfrac{x^2-\left(x^2-1\right)}{x\left(x+1\right)}}=\dfrac{\dfrac{1}{x\left(x-1\right)}}{\dfrac{1}{x\left(x+1\right)}}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{0;\pm1\right\}\\A=\dfrac{x+1}{x-1}\end{matrix}\right.\)
a: \(=\dfrac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{\left(2a-1\right)\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}-\dfrac{6a^2+6a+1}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{4a^2-3a+5+2a^2-3a+1-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{-12a}{\left(a-1\right)\left(a^2+a+1\right)}\)
b: \(=\dfrac{5}{a+1}+\dfrac{10}{a^2-a+1}-\dfrac{15}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\dfrac{5a^2-5a+5+10a+10-15}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\dfrac{5a^2+5a}{\left(a+1\right)\left(a^2-a+1\right)}=\dfrac{5a}{a^2-a+1}\)
a.
\(\dfrac{\dfrac{1}{x}+\dfrac{1}{y}}{\dfrac{1}{x}-\dfrac{1}{y}}=\dfrac{\dfrac{x+y}{xy}}{\dfrac{y-x}{xy}}=\dfrac{x+y}{y-x}\)
b.
\(\dfrac{\dfrac{x}{x+1}-\dfrac{x-1}{x}}{\dfrac{x}{x-1}-\dfrac{x+1}{x}}=\dfrac{\dfrac{x^2-\left(x+1\right)\left(x-1\right)}{x\left(x+1\right)}}{\dfrac{x^2-\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}}=\dfrac{\dfrac{1}{x\left(x+1\right)}}{\dfrac{1}{x\left(x-1\right)}}=\dfrac{x\left(x+1\right)}{x\left(x-1\right)}=\dfrac{x+1}{x-1}\)
c.
\(1-\dfrac{x}{1-\dfrac{x}{x+1}}=1-\dfrac{x}{\dfrac{1}{x+1}}=1-\dfrac{x+1}{x}=\dfrac{x-\left(x+1\right)}{x}=\dfrac{-1}{x}\)
d.
