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a: =>(x^2-3x)^2+3(x^2-3x)=0

=>(x^2-3x)(x^2-3x+3)=0

=>x=0 hoặc x=3

b: Đặt x^2-5=a

=>\(a+\sqrt{a-1}=7\)

=>a-1+căn a-1-6=0

=>(căn a-1+3)(căn a-1-2)=0

=>căn a-1=2

=>a-1=4

=>a=5

=>x^2-5=5

=>x^2=10

=>\(x=\pm\sqrt{10}\)

\(\sqrt{\dfrac{4}{9}}:\sqrt{\dfrac{25}{36}}\)

\(=\dfrac{\sqrt{4}}{\sqrt{9}}:\dfrac{\sqrt{25}}{\sqrt{36}}\)

\(=\dfrac{\sqrt{2^2}}{\sqrt{3^2}}:\dfrac{\sqrt{5^2}}{\sqrt{6^2}}\)

\(=\dfrac{2}{3}:\dfrac{5}{6}\)

\(=\dfrac{2}{3}\cdot\dfrac{6}{5}\)

\(=\dfrac{4}{5}\)

\(\dfrac{\sqrt{80}}{\sqrt{5}}\)

\(=\dfrac{\sqrt{5\cdot16}}{\sqrt{5}}\)

\(=\sqrt{\dfrac{5\cdot16}{5}}\)

\(=\sqrt{16}\)

\(=\sqrt{4^2}\)

\(=4\)

a: \(x^2=2\)

=>\(x^2=\left(\sqrt{2}\right)^2\)

=>\(x=\pm\sqrt{2}\)

b: \(x^2=9\)

=>\(x^2=3^2\)

=>\(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

c: \(\left(x-\sqrt{2}\right)^2=2\)

=>\(\left[{}\begin{matrix}x-\sqrt{2}=\sqrt{2}\\x-\sqrt{2}=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}\\x=0\end{matrix}\right.\)

d: \(4x^2-1=0\)

=>\(4x^2=1\)

=>\(x^2=\dfrac{1}{4}\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)