\(\frac{9}{16}x^2-\frac{1}{4}=0\)
\(\Leftrightarrow\left(\frac{3}{4}x\right)^2-\left(\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(\frac{3}{4}x-\frac{1}{2}\right)\left(\frac{3}{4}x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\frac{3}{4}x-\frac{1}{2}=0\\\frac{3}{4}x+\frac{1}{2}=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{2}{3}\\x=-\frac{2}{3}\end{array}\right.\)
Bài này cũng có cách khác mà
\(\frac{9}{16}x^2-\frac{1}{4}=0\)
\(\left(\frac{3}{4}x\right)^2-\left(\frac{1}{2}\right)^2=0\)
\(\left(\frac{3}{4}x-\frac{1}{2}\right)\left(\frac{3}{4}x+\frac{1}{2}\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}\frac{3}{4}x-\frac{1}{2}=0\\\frac{3}{4}x+\frac{1}{2}=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}\frac{3}{4}x=\frac{1}{2}\\\frac{3}{4}x=-\frac{1}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{2}{3}\\x=-\frac{2}{3}\end{array}\right.\)
\(\frac{9}{16}x^2-\frac{1}{4}=0\)
\(\frac{9}{16}x^2=\frac{1}{4}\)
\(x^2=\frac{4}{9}\)
\(x^2=\left(\frac{2}{3}\right)^2=\left(-\frac{2}{3}\right)^2\)
Vậy \(x=\frac{2}{3};-\frac{2}{3}\)
