`a)` `\sqrt{x^2-10x+25}+\sqrt{4x^2+4x+1}`
`=\sqrt{(x-5)^2}+\sqrt{(2x+1)^2}`
`=|x-5|+|2x+1|`
Vậy: `\sqrt{x^2-10x+25}+\sqrt{4x^2+4x+1}`
`=|x-5|+|2x+1|`
$\\$
`b)` `3|x-1/2|-2\sqrt{x^2+2x+1}-1/3`
`=3|x-1/2|-2\sqrt{(x+1)^2}-1/3`
`=3|x-1/2|-2|x+1|-1/3`
Vậy: `3|x-1/2|-2\sqrt{x^2+2x+1}-1/3`
`=3|x-1/2|-2|x+1|-1/3`
$\\$
`c)` `{\sqrt{x-1}(\sqrt{x}-2)}/\sqrt{x-4\sqrt{x}+4}` `(x\ge 1; x\ne 4)`
`={\sqrt{x-1}(\sqrt{x}-2)}/\sqrt{(\sqrt{x}-2)^2}`
`={\sqrt{x-1}(\sqrt{x}-2)}/|\sqrt{x}-2|`
+) Nếu `x>4=>\sqrt{x}>=2`
`=>|\sqrt{x}-2|=\sqrt{x}-2`
`=>{\sqrt{x-1}(\sqrt{x}-2)}/|\sqrt{x}-2|`
`={\sqrt{x-1}(\sqrt{x}-2)}/{\sqrt{x}-2}`
`=\sqrt{x-1}`
$\\$
+) Nếu `1\le x<4=>\sqrt{x}<2`
`=>{\sqrt{x-1}(\sqrt{x}-2)}/|\sqrt{x}-2|`
`={\sqrt{x-1}(\sqrt{x}-2)}/{-(\sqrt{x}-2)}`
`=-\sqrt{x-1}`
Vậy:
$\dfrac{\sqrt{x-1}(\sqrt{x}-2)}{\sqrt{x-4\sqrt{x}+4}}=\begin{cases}\sqrt{x-1}\ nếu \ x>4\\-\sqrt{x-1}\ nếu \ 1\le x<4\end{cases}$