a) Ta có: \(x^2-25=0\)
\(\Leftrightarrow x^2=25\)
Vậy \(x=\sqrt{25}=-5;5\)
b) Ta có: \(\dfrac{1}{3}x-2=0\)
\(\Leftrightarrow\dfrac{1}{3}x=2\)
Vậy \(x=\dfrac{2}{\dfrac{1}{3}}=6\)
c)Theo đề bài ta có: \(\left(2x+\dfrac{1}{3}\right)\left(x-\dfrac{2}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{1}{3}=0\\x-\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(\dfrac{1}{6};\dfrac{2}{3}\right)\)
d)Theo đề bài, ta có: \(10x^2-8x=0\)
\(\Rightarrow x\left(10x-8\right)=0\)
\(\left[{}\begin{matrix}x=0\\10x-8=0\Rightarrow x=\dfrac{4}{5}\end{matrix}\right.\)
Vậy x = 0 hoặc x = \(\dfrac{4}{5}\)
e) Ta có: \(x^2-5x+4=0\)
\(\Leftrightarrow x^2-4x-x+4=0\)
\(\Leftrightarrow x\left(x-4\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
Vậy x = 1;4
