Lê Quốc Trần Anh,you mustn't deduce \(x=\sqrt{a}\) from \(x^2=a\left(a\ge0\right)\)since x has 2 values : \(\pm\sqrt{a}\)

d) \(x=\pm\sqrt{5,76}=\pm2,4\)

e) \(\left[{}\begin{matrix}x-2=-6\\x-2=6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=8\end{matrix}\right.\)

Lê Quốc Trần Anh :v , you wrong 

e) \(\dfrac{x-2}{9}=\dfrac{4}{x-2}\)

\(\Leftrightarrow\left(x-2\right)^2=36\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-4\end{matrix}\right.\)

\(\dfrac{x-2}{9}=\dfrac{4}{x-2}\)

<=> \(\left(x-2\right)\left(x-2\right)=4.9\)

<=> \(\left(x-2\right)^2=36\)

<=> \(x-2=\sqrt{36}\)

<=> \(x-2=6\) <=> \(x=6-2=4\)

b, 2,5:x=1,5:12

    2,5:x=0,125

    x = 2,5:0,125

    x = 20

d, \(\dfrac{x}{6}=\dfrac{24}{25x}\) => \(x.25x=6.24\) => \(x^2.25=144\) <=> \(x^2=144:25=5,76\)<=> \(x=\sqrt{5,76}=2,4\)