a, x mũ 2 = 16/49 b, x mũ 2 = 36/49 c, ( x + 1/2 ) mũ 2 = 9/4 d, ( x – 2/3 ) mũ 2 = 1/8 Giúp đi nha

`x^2=16/49`

\(⇔\left[\begin{array}{l}x^2=(\dfrac{4}{7})^2\\x^2=(-\dfrac{4}{7})^2\end{array} \right.\)

\(⇔\left[\begin{array}{l}x=\dfrac{4}{7}\\x=-\dfrac{4}{7}\end{array} \right.\)

`x^2=36/49`

\(⇔\left[\begin{array}{l}x^2=(\dfrac{6}{7})^2\\x^2=(-\dfrac{6}{7})^2\end{array} \right.\)

\(⇔\left[\begin{array}{l}x=\dfrac{6}{7}\\x=-\dfrac{4}{7}\end{array} \right.\)

`(x+1/2)^2=9/4`

\(⇔\left[\begin{array}{l}x+\dfrac{1}{2}=\dfrac{3}{2}\\x+\dfrac{1}{2}=-\dfrac{3}{2}\end{array} \right.\)

\(⇔\left[\begin{array}{l}x=1\\x=-2\end{array} \right.\)

`(x-2/3)^2=1/8`

\(⇔\left[\begin{array}{l}x-\dfrac{2}{3}=2\sqrt{2}\\x-\dfrac{2}{3}=-2\sqrt{2}\end{array} \right.\)

\(⇔\left[\begin{array}{l}x=\dfrac{2+6\sqrt{2}}{3}\\x=\dfrac{2-6\sqrt{2}}{3}\end{array} \right.\)