[tex]\bf \cfrac{2x^2-5}{x^2-4}-1=\cfrac{x^2+7}{x^2+1}\impliedby \textit{let's add the left-side}
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\cfrac{2x^2-5-x^2+4}{x^2-4}=\cfrac{x^2+7}{x^2+1}\implies \cfrac{x^2-1}{x^2-4}=\cfrac{x^2+7}{x^2+1}
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\textit{let's cross-multiply now}
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(x^2-1)(x^2+1)=(x^2+7)(x^2-4)
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x^4-1=x^4+3x^2-28\implies 0=3x^2-27
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27=3x^2\implies \cfrac{27}{3}=x^2\implies 9=x^2\implies \pm 3=x[/tex]
restrictions... well, for a fraction, the restrictions will be, that a denominator doesn't become 0, otherwise, the fraction becomes a division by 0, which is an "undefined" value
so.... you have two fractions in the original expression
if we set those denominators to 0, then we can see what "x" is when that expression is 0, and thus, those values of "x", will make the denominator 0, and thus, are not values you'd want in the "domain", because those values make the fraction undefined
so [tex]\bf \begin{cases}
x^2-4=0\implies x^2=4\implies x=\pm 2\\\\
x^2+1=0\implies x^2=-1\implies x=\pm i
\end{cases}[/tex]
the second denominator, gives a +/- i, is just another way to say, the expression will never be 0 no matter what you give "x"
so, the only restriction will then be, from the first denominator, "x" can be anything BUT +/- 2
because if "x" ever becomes 2 or -2, that denominator goes to 0, and the fraction will go poof
