\(c,ĐK:x\le\dfrac{1}{3}\\ =5x-\dfrac{\left|3x-1\right|}{1-3x}=\left[{}\begin{matrix}5x-\dfrac{3x-1}{1-3x}\left(x\ge\dfrac{1}{3}\right)\\5x-\dfrac{1-3x}{1-3x}\left(x< \dfrac{1}{3}\right)\end{matrix}\right.=\left[{}\begin{matrix}5x+1\left(x\ge\dfrac{1}{3}\right)\\5x-1\left(x< \dfrac{1}{3}\right)\end{matrix}\right.\)
\(e,=\left(2+\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\left(2-\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(a\ge0;a\ne1\right)\\ =\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a\\ f,=\dfrac{x+\sqrt{x}+\sqrt{x}-x}{\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\left(x\ge0;x\ne1\right)\\ =\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
c) \(ĐK:x\ne\dfrac{1}{3}\)
\(=5x-\dfrac{\sqrt{\left(1-3x\right)^2}}{1-3x}=5x-\dfrac{\left|1-3x\right|}{1-3x}=\) \(\left[{}\begin{matrix}5x-\dfrac{1-3x}{1-3x}=5x-1\left(x< \dfrac{1}{3}\right)\\5x-\dfrac{3x-1}{1-3x}=5x+1\left(x>\dfrac{1}{3}\right)\end{matrix}\right.\)
e) \(ĐK:a\ge0,a\ne1\)
\(=\left(2+\dfrac{\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\right)\left(2-\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a\)
f) \(ĐK:x>0,x\ne1\)
\(=\dfrac{\sqrt{x}+x+\sqrt{x}-x}{1-x}.\dfrac{1-x}{-\sqrt{x}}=\dfrac{2\sqrt{x}}{-\sqrt{x}}=-2\)
