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a.
\(=\sqrt{\sqrt{5}-2}-\sqrt{5\left(\sqrt{5}+2\right)}+2\sqrt{\sqrt{5}+2}\)
\(=\sqrt{\sqrt{5}-2}-\sqrt{\sqrt{5}+2}\left(\sqrt{5}-2\right)\)
\(=\sqrt{\sqrt{5}-2}-\sqrt{\sqrt{5}-2}\left(\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\right)\)
\(=\sqrt{\sqrt{5}-2}-\sqrt{\sqrt{5}-2}.1=0\)
b.
\(=\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}-\sqrt{2\left(\sqrt{2}+1\right)}\)
\(=\sqrt{\sqrt{2}-1}-\left(\sqrt{2}-1\right)\left(\sqrt{\sqrt{2}+1}\right)\)
\(=\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}-1}.\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
\(=\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}-1}=0\)
b: \(=\dfrac{\sqrt{5}+1}{\sqrt{5}-1}+\dfrac{\sqrt{5}-1}{\sqrt{5}+1}\)
\(=\dfrac{6+2\sqrt{5}+6-2\sqrt{5}}{4}=\dfrac{12}{4}=3\)
c: \(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\)
e: \(=\dfrac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{\sqrt{2}\cdot\sqrt{3+\sqrt{3}-1}}{\sqrt{3}-1}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}-1}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(=\dfrac{4-2\sqrt{3}}{2}=2-\sqrt{3}\)
c) \(\sqrt{5+\sqrt{24}}=\sqrt{5+2\sqrt{6}}=\sqrt{3}+\sqrt{2}\)
d) \(\sqrt{12-\sqrt{140}}=\sqrt{12-2\sqrt{35}}=\sqrt{7}-\sqrt{5}\)
f) \(\sqrt{8-\sqrt{28}}=\sqrt{8-2\sqrt{7}}=\sqrt{7}-1\)
g) \(\sqrt{23-4\sqrt{15}}=\sqrt{23-2\cdot\sqrt{60}}=2\sqrt{5}-\sqrt{3}\)
h) \(\sqrt{9+4\sqrt{2}}=\sqrt{\left(2\sqrt{2}+1\right)^2}=2\sqrt{2}+1\)
\(\sqrt{8-4\sqrt{3}}-\sqrt{8+4\sqrt{3}}=\sqrt{5-4\sqrt{3}+3}-\sqrt{5+4\sqrt{3}+3}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}=-2\sqrt{3}\)
\(\sqrt{9-\sqrt{5\sqrt{3}+5\sqrt{8+10\sqrt{7-4\sqrt{3}}}}}=\sqrt{9-\sqrt{5\sqrt{3}+5\sqrt{8+10\sqrt{4-4\sqrt{3}+3}}}}=\sqrt{9-\sqrt{5\sqrt{3}+5\sqrt{8+10\sqrt{\left(2-\sqrt{3}\right)^2}}}}=\sqrt{9-\sqrt{5\sqrt{3}+5\sqrt{8+10\left(2-\sqrt{3}\right)}}}=\sqrt{9-\sqrt{5\sqrt{3}+5\sqrt{8}+20-10\sqrt{3}}}=\sqrt{9-\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}=\sqrt{9-\sqrt{5\sqrt{3}+5\sqrt{25-10\sqrt{3}}+3}}=\sqrt{9-\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}=\sqrt{9-\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}=\sqrt{9-\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{9-\sqrt{25}}=\sqrt{9-5}=\sqrt{4}=2\)
c.
(\sqrt{5}-\sqrt{3})-(\sqrt{10}-\sqrt{7})=(\sqrt{5}+\sqrt{7})-(\sqrt{3}+\sqrt{10})
Mà:
\((\sqrt{5}+\sqrt{7})^2=12+\sqrt{35}< 12+\sqrt{36}=18\)
\((\sqrt{3}+\sqrt{10})^2=13+\sqrt{30}>13+\sqrt{25}=18\)
\(\Rightarrow \sqrt{3}+\sqrt{10}> \sqrt{5}+\sqrt{7}\Rightarrow \sqrt{5}-\sqrt{3}< \sqrt{10}-\sqrt{7}\)
Lời giải:
a.
$5+\sqrt{2}>5+\sqrt{1}=6$
$4+\sqrt{3}< 4+\sqrt{4}=6$
$\Rightarrow 5+\sqrt{2}>4+\sqrt{3}$
b.
$\sqrt{8}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2}$
$\sqrt{5}-\sqrt{3}=\frac{5-3}{\sqrt{5}+\sqrt{3}}=\frac{2}{\sqrt{5}+\sqrt{3}}< \frac{2}{\sqrt{2}}=\sqrt{2}$
Vậy $\sqrt{8}-\sqrt{2}>\sqrt{5}-\sqrt{2}$