Write the solution set of
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Part I : ![]()
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Square term is always positive and (x + 1)2 ≠ 0. So, x > 0 and x ≠ -1
Part II : ![]()
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Square term is always positive and (x – 1)2 ≠ 0. So, x < 0 and x ≠ 1
So, by taking union of part I and part II we have final solution i.e., x ∈ R – {-1, 0, 1}