Question

tim x biet

5-9\(x^2=0\)

\(x^2+x+\dfrac{1}{4}=0\)

Answer 1

\(5-9x^2=0\)

\(\Leftrightarrow9x^2=5\)

\(\Leftrightarrow x^2=\dfrac{5}{9}\)

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{5}{9}}\\x=-\sqrt{\dfrac{5}{9}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}}{3}\\x=-\dfrac{\sqrt{5}}{3}\end{matrix}\right.\)

\(x^2+x+\dfrac{1}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=0\)

\(\Rightarrow x+\dfrac{1}{2}=0\Rightarrow x=-\dfrac{1}{2}\)

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Answer 2

\(5-9x^2=0\\ 9x^2=5\\ x^2=\dfrac{5}{9}\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{-\sqrt{5}}{3}\\x=\dfrac{\sqrt{5}}{3}\end{matrix}\right.\)

\(x^2+x+\dfrac{1}{4}=0\\ \left(x+\dfrac{1}{2}\right)^2=0\\ x+\dfrac{1}{2}=0\\ x=\dfrac{-1}{2}\)