Đáp án:
`a)` $ĐKXĐ:$ `x\ne +-1/2; x\ne0`
`A=(1/(2x-1)-2/(2x+1)+3/(1-4x^2)):(8x^2)/(4x^2-1)`
`A=(1/(2x-1)-2/(2x+1)-3/(4x^2-1)):(8x^2)/((2x-1)(2x+1))`
`A=(1/(2x-1)-2/(2x+1)-3/((2x-1)(2x+1)). ((2x-1)(2x+1))/(8x^2)`
`A=((2x+1)/((2x-1)(2x+1))-(2.(2x-1))/((2x-1)(2x+1))-3/((2x-1)(2x+1))). ((2x-1)(2x+1))/(8x^2)`
`A=(2x+1-4x+2-3)/((2x-1)(2x+1)). ((2x-1)(2x+1))/(8x^2)`
`A=(-2x)/((2x-1)(2x+1)). ((2x-1)(2x+1))/(8x^2)`
`A=(-2x)/(8x^2)=(-1)/(4x)`
`b)` `(x^2-5).A=1`
`<=> (x^2-5).((-1)/(4x))=1`
`<=> x^2-5=-4x`
`=> x^2-5+4x=0`
`=> x^2-x+5x-5=0`
`=> x(x-1)+5(x-1)=0`
`=> (x-1)(x+5)=0`
`=>` \(\left[ \begin{array}{l}x-1=0\\x+5=0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=1\\x=-5\end{array} \right.\)
