\(b.MCNN:12x\left(x-1\right)\left(x+1\right)\)

\(\dfrac{5}{4x-4}=\dfrac{5}{4\left(x-1\right)}=\dfrac{15x\left(x+1\right)}{12x\left(x-1\right)\left(x+1\right)}=\dfrac{15x^2+15x}{12\left(x-1\right)\left(x+1\right)};\dfrac{4x}{1-x^2}=-\dfrac{4x.12x}{12x\left(x-1\right)\left(x+1\right)}=-\dfrac{48x^2}{12\left(x-1\right)\left(x+1\right)};\dfrac{1}{3x^2+3x}=\dfrac{1}{3x\left(x+1\right)}=\dfrac{4\left(x-1\right)}{12x\left(x-1\right)\left(x+1\right)}\)

\(a.MCNN:\left(3x-1\right)\left(3x+1\right)\)

\(\dfrac{2}{9x^2-1}=\dfrac{2}{\left(3x-1\right)\left(3x+1\right)};\dfrac{4x}{3x-1}=\dfrac{4x\left(3x+1\right)}{\left(3x-1\right)\left(3x+1\right)}\)

a)\(\dfrac{2x-10}{x^2-25}=\dfrac{2\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{2}{x+5}\)

b)\(\dfrac{4x^2-4x+1}{2-4x}=\dfrac{\left(2x-1\right)^2}{-2\left(2x-1\right)}=-\dfrac{2x-1}{2}\)

c)\(\dfrac{16a^2-1}{16a^2-8a+1}=\dfrac{\left(4a-1\right)\left(4a+1\right)}{\left(4a-1\right)^2}=\dfrac{4a+1}{4a-1}\)

d)\(\dfrac{x^3+1}{x^2-1}=\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-x+1}{x-1}\)

\(b.\sqrt{x^2-6x+9}=5\)

\(\Leftrightarrow\)\(\sqrt{\left(x-3\right)^2}=5\)

\(\Leftrightarrow\)\(\left|x-3\right|=5\)\(\Rightarrow\)Với x \(\ge3\)\(\Rightarrow\) \(x=8\)

                         Với x \(< 3=>x=-2\)

\(Để\) N có nghĩa thì : \(\left\{{}\begin{matrix}\sqrt{x-6}\ge0\\\sqrt{x-6}\ne0\end{matrix}\right.=>\sqrt{x-6}>0=>x-6>0=>x>6\)

Vậy để N có nghĩa thì x > 6