Z definicji wartości bezwzględnej:

$\left|x\right|=\{\begin{array}{l}x\text{dla}x\ge 0\\ -x\text{dla}x<0\end{array}$  

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$\text{a)}\left|\sqrt{2}+\sqrt{3}-2\right|=\sqrt{2}+\sqrt{3}-2$  

Uzasadnienie:

$\sqrt{2}+\sqrt{3}>\sqrt{1}+\sqrt{1}=1+1=2\Rightarrow \sqrt{2}+\sqrt{3}-2>0$  

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$\text{b)}\left|5-2\sqrt{3}-\sqrt{5}\right|=-\left(5-2\sqrt{3}-\sqrt{5}\right)=-5+2\sqrt{3}+\sqrt{5}=2\sqrt{3}+\sqrt{5}-5$  

Uzasadnienie:

$2\sqrt{3}+\sqrt{5}>2\sqrt{2,25}+\sqrt{4}=2\cdot 1,5+2=3+2=5\Rightarrow 5-\left(2\sqrt{3}+\sqrt{5}\right)<0$  

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$\text{c)}\left|\sqrt{10}+\sqrt{17}-\sqrt{47}\right|=\sqrt{10}+\sqrt{17}-\sqrt{47}$  

Uzasadnienie:

$\sqrt{47}<\sqrt{49}=7$  

$\sqrt{10}+\sqrt{17}>\sqrt{9}+\sqrt{16}=3+4=7$  

$\sqrt{10}+\sqrt{17}-\sqrt{47}>0$  

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$\text{d)}\left|\sqrt{99}-3-\sqrt{50}\right|=-\left(\sqrt{99}-3-\sqrt{50}\right)=-\sqrt{99}+3+\sqrt{50}=3+\sqrt{50}-\sqrt{99}$  

Uzasadnienie:

$\sqrt{99}<\sqrt{100}=10$  

$3+\sqrt{50}>3+\sqrt{49}=3+7=10$  

$\sqrt{99}-\left(3+\sqrt{50}\right)<0$