a, $9x^2=25\Rightarrow9x^2=\left(\sqrt{25}\right)^2$

$\Rightarrow9x=\pm\sqrt{25}\Rightarrow9x=\pm5$

$\Rightarrow x=\pm\dfrac{5}{9}$

b, $x^3-\dfrac{1}{4}x=0\Rightarrow x^2\left(x-\dfrac{1}{4}\right)=0$

$\Rightarrow\left[{}\begin{matrix}x^2=0\\x-\dfrac{1}{4}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.$

c, $9x^2-36=0\Rightarrow9x^2=36$

$\Rightarrow9x^2=6^2\Rightarrow9x=6$

$\Rightarrow x=\dfrac{2}{3}$

x(x-1)-3x+x=0

Ta có :

$x\left(x-1\right)-3x+x=0$

$\Rightarrow x^2-x-3x+x=0$

$\Rightarrow x^2-3x=0$

$\Rightarrow x\left(x-3\right)=0$

$\Rightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.$

Vậy $\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.$

tìm x biết

( x^2 -25)^2 -(x-5)^2 =0

$\left(x^2-25\right)^2-\left(x-5\right)^2=0$

$\Leftrightarrow\left[\left(x-5\right)\left(x+5\right)\right]^2-\left(x-5\right)^2=0$

$\Leftrightarrow\left(x-5\right)^2\cdot\left(x+5\right)^2-\left(x-5\right)^2=0$

$\Leftrightarrow\left(x-5\right)^2\cdot\left[\left(x+5\right)^2-1\right]=0$

$\Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)^2=0\\\left(x+5\right)^2-1=0\Rightarrow\left(x+5\right)^2=1\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+5=1\\x+5=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\\x=-6\end{matrix}\right.$

Vậy: x= 5; x=-4 hoặc x=-6