Question

\(x^2-5=\left(2x-\sqrt{5}\right)\cdot\left(x+\sqrt{5}\right)\)

Answer 1

\(x^2-5=\left(2x-\sqrt{5}\right).\left(x+\sqrt{5}\right)\)

\(\Leftrightarrow\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)-\left(2x-\sqrt{5}\right).\left(x+\sqrt{5}\right)=0\)

\(\Leftrightarrow\left(x+\sqrt{5}\right)\left(x-\sqrt{5}-2x+\sqrt{5}\right)=0\)

\(\Leftrightarrow\left(x+\sqrt{5}\right).\left(-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{5}=0\\-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{5}\\x=0\end{matrix}\right.\)

Vậy PT có nghiệm là \(x=0;x=-\sqrt{5}\)

P/S : chắc là đúng :D