\(\text{a) }\sqrt{9-2\sqrt{14}}=\sqrt{7+2-2\sqrt{14}}\\ =\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}=\sqrt{7}-\sqrt{2}\)
\(\text{b) }\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\\ \left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)^2\\ =4-\sqrt{7}-2\sqrt{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}+4+\sqrt{7}\\ =8-2\sqrt{16-7}\\ =8-2\sqrt{9}=8-6=2\\ \Rightarrow\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=\sqrt{2}\)
\(\text{c) }\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\\ \left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)^2\\ =4+\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}+4-\sqrt{10+2\sqrt{5}}\\ =8+2\sqrt{16-10-2\sqrt{5}}\\ =8+2\sqrt{5+1-2\sqrt{5}}\\ =8+2\sqrt{\left(\sqrt{5}-1\right)^2}\\ =8+2\sqrt{5}-2\\ =5+1+2\sqrt{5}\\ =\left(\sqrt{5}+1\right)^2\\ \Rightarrow\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}=\sqrt{5}+1\)
a) \(\sqrt{9-2\sqrt{14}}=\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}=\left|\sqrt{7}-\sqrt{2}\right|=\sqrt{7}-\sqrt{2}\)
b) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=\dfrac{\sqrt{2}\sqrt{4-\sqrt{7}}}{\sqrt{2}}-\dfrac{\sqrt{2}\sqrt{4+\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{2\left(4-\sqrt{7}\right)}}{\sqrt{2}}-\dfrac{\sqrt{2\left(4+\sqrt{7}\right)}}{\sqrt{2}}=\dfrac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}-\dfrac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}-\dfrac{\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}=\dfrac{\left|\sqrt{7}-1\right|}{\sqrt{2}}-\dfrac{\left|\sqrt{7}+1\right|}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}-1}{\sqrt{2}}-\dfrac{\sqrt{7}+1}{\sqrt{2}}=\dfrac{\sqrt{7}-1-\left(\sqrt{7}+1\right)}{\sqrt{2}}=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}\)
\(=\dfrac{-2}{\sqrt{2}}=\dfrac{-\sqrt{2}\sqrt{2}}{\sqrt{2}}=-\sqrt{2}\)
