a.
\(\left(3x-5\right)^2=\dfrac{25}{9}\)
\(\Rightarrow\left[{}\begin{matrix}3x-5=\sqrt{\dfrac{25}{9}}\\3x-5=-\sqrt{\dfrac{25}{9}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x-5=\dfrac{5}{3}\\3x-5=-\dfrac{5}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{20}{9}\\x=\dfrac{10}{9}\end{matrix}\right.\)
Vậy:............
b.
\(\left\{{}\begin{matrix}\left(x-2\right)^{2018}\ge0\\\left|y^2-9\right|^{2014}\ge0\\\left(x-2\right)^{2018}+\left|y^2-9\right|^{2014}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^{2018}=0\\\left|y^2-9\right|^{2014}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=\pm3\end{matrix}\right.\)
Tìm x, biết:
a. (3x -5)2 = \(\left(\dfrac{25}{9}\right)\)
(3x -5)2 = \(\left(\dfrac{5}{3}\right)^2\)
=> 3x - 5 = \(\dfrac{5}{3}\)
3x = \(\dfrac{5}{3}\) + 5
3x = \(\dfrac{14}{3}\)
x = \(\dfrac{14}{3}\) : 3
x = \(\dfrac{14}{9}\)
a)
\(\left(3x-5\right)^2=\dfrac{25}{9}\\ \Rightarrow\left[{}\begin{matrix}3x-5=\dfrac{5}{3}\\3x-5=-\dfrac{5}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{20}{9}\\x=\dfrac{10}{9}\end{matrix}\right.\)
b)
\(\left(x-2\right)^{2018}\ge0\\ \left|y^2-9\right|^{2014}\ge0\\ \left(x-2\right)^{2018}+\left|y^2-9\right|^{2014}=0\\ \Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^{2018}=0\\\left|y^2-9\right|^{2014}=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=2\\y=\pm3\end{matrix}\right.\)
