\(36x^2-49=0\)
\(\Leftrightarrow36x^2=49\)
\(\Leftrightarrow x=\sqrt{\dfrac{49}{36}}=\dfrac{7}{6}\)
Vậy: \(x=\dfrac{7}{6}\)
\((6x)^2-7^2=0\)
(6x-7)(6x+7)=0
Th1 6x-7=0
X=7/6
Th2 6x+7=0
X=-7/6
Pt có tập nghiệm S=7/6;-7/6
Ta có: \(36x^2-49=0\)
\(\Leftrightarrow\left(6x-7\right)\left(6x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)
a) \(36x^2-49=0\Rightarrow\left(6x-7\right)\left(6x+7\right)=0\Rightarrow\left[{}\begin{matrix}6x-7=0\\6x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)
\(36x^2-49=0\\ \Leftrightarrow36x^2=49\\ \Leftrightarrow x^2=\dfrac{49}{36}\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)
Vậy...................
\(\)\(\left(6x\right)^2-7^2=0\Rightarrow\left(6x-7\right).\left(6x+7\right)=0\Rightarrow\left[{}\begin{matrix}6x-7=0\\6x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6x=7\\6x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)
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