x2 - 2x - 4 = 0
⇔ ( x2 - 2x + 1 ) - 5 = 0
⇔ ( x - 1 )2 - ( √5 )2 = 0
⇔ ( x - 1 - √5 )( x - 1 + √5 ) = 0
⇔ x = √5 + 1 hoặc x = -√5 + 1
\(x^2-2x-4=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow\left(x-1\right)^2-5=0\Leftrightarrow\left(x-1-\sqrt{5}\right)\left(x-1+\sqrt{5}\right)=0\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
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