Question

Tìm x:

x² + 5x = \(\sqrt{37}\)

Answer 1

\(x^2+5x=\sqrt{37}\)

\(\Leftrightarrow4x^2+20x=4\sqrt{37}\)

\(\Leftrightarrow4x^2+20x+25=4\sqrt{37}+25\)

\(\Leftrightarrow\left(2x+5\right)^2=4\sqrt{37}+25\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+5=\sqrt{4\sqrt{37}+25}\\2x+5=-\sqrt{4\sqrt{37}+25}\end{matrix}\right.\Leftrightarrow x=\dfrac{\pm\sqrt{4\sqrt{37}+25}-5}{2}\)

Answer 2

\(x^2+5x=\sqrt{37}\)

\(\Leftrightarrow x\left(x+5\right)=\sqrt{37}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{37}\\x+5=\sqrt{37}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{37}\\x=-5+\sqrt{37}\end{matrix}\right.\)

Vậy \(x=\sqrt{37};x=-5+\sqrt{37}\).