Ta có:
\(\left|x^2-5\right|=\left|x-\sqrt{5}\right|\)
\(\Leftrightarrow\left|\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\right|=\left|x-\sqrt{5}\right|\)
\(\Leftrightarrow\left|x-\sqrt{5}\right|.\left|x+\sqrt{5}\right|=\left|x-\sqrt{5}\right|\) (*)
+) Nếu \(\left|x-\sqrt{5}\right|=0\) thì \(x=\sqrt{5}\)
+) Nếu \(\left|x-\sqrt{5}\right|\ne0\) thì phương trình (*) trở thành:
\(\left|x+\sqrt{5}\right|=1\)
\(\Leftrightarrow x=\pm1-\sqrt{5}\)
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