\(\sqrt{x+y}+\sqrt{x^2-y^2}\\ =\sqrt{x+y}+\sqrt{\left(x+y\right)\left(x-y\right)}\\ =\sqrt{x+y}+\sqrt{x+y}.\sqrt{x-y}\\ =\sqrt{x+y}\left(1+\sqrt{x-y}\right)\)

\(\sqrt{5-2\sqrt{6}}-\sqrt{5+2\sqrt{6}}\\ =\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\\ =\left|\sqrt{3}-\sqrt{2}\right|-\left|\sqrt{3}+\sqrt{2}\right|\\ =\sqrt{3}-\sqrt{2}-\sqrt{3}-\sqrt{2}\\ =-2\sqrt{2}\)

+)Phép chia hết: \(a:b=x\) với \(b\ne0\) thì \(a=b\times x\)

+)Phép chia có dư: \(a:b=y\) (dư \(r\)) với \(b\ne0;r< b\) thì \(a=b\times y+r\)

\(x^2\left(x-1\right)-16\left(x-1\right)=0\\ \left(x^2-16\right)\left(x-1\right)=0\\ \left(x-4\right)\left(x+4\right)\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\\x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=1\end{matrix}\right.\)

Vậy..........

\(36x^2-49=0\\ \Leftrightarrow36x^2=49\\ \Leftrightarrow x^2=\dfrac{49}{36}\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)

Vậy...................