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a: Ta có: \(\sqrt{\left(5-\sqrt{19}\right)^2}-\sqrt{\left(4-\sqrt{19}\right)^2}\)
\(=5-\sqrt{19}-\sqrt{19}+4\)
\(=9-2\sqrt{19}\)
b: Ta có: \(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(2\sqrt{2}-3\right)^2}\)
\(=3-2\sqrt{2}-3+2\sqrt{2}\)
=0
c.
Căn bậc 2 không xác định do $2-\sqrt{5}< 0$
d.
\(=\sqrt{(3+\sqrt{3})^2}(3+\sqrt{3})=|3+\sqrt{3}|(3+\sqrt{3})=(3+\sqrt{3})^2=12+6\sqrt{3}\)
e.
\(=(2-\sqrt{5})\sqrt{(2+\sqrt{5})^2}=(2-\sqrt{5})|2+\sqrt{5}|=(2-\sqrt{5})(2+\sqrt{5})=4-5=-1\)
a,\(\left(\sqrt{6}+\sqrt{2}\right)\sqrt{2-\sqrt{3}}\)
\(=\sqrt{2}\left(\sqrt{3}+1\right)\sqrt{2-\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)
\(=3-1\)
\(=2\)
b, \(\left(\sqrt{4+\sqrt{15}}-\sqrt{16-3\sqrt{15}}\right)\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{8+2\sqrt{15}}-\sqrt{32-6\sqrt{15}}}{\sqrt{2}}.\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{3+2\sqrt{3}.\sqrt{5}+5}-\sqrt{27-2.3\sqrt{3}.\sqrt{5}+5}}{\sqrt{2}}\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{3}-\sqrt{5}\right)^2}}{\sqrt{2}}\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{3}+\sqrt{5}-3\sqrt{3}+\sqrt{5}}{\sqrt{2}}\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{2\sqrt{5}-2\sqrt{3}}{\sqrt{2}}\left(\sqrt{5}+\sqrt{3}\right)\)
\(=\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)\)
\(=\sqrt{2}\left(5-3\right)\)
\(=2\sqrt{2}\)
\(A=\dfrac{\sqrt{2+\sqrt{4-x^2}}\left(\sqrt{\left(2+x\right)^3}-\sqrt{\left(2-x\right)^3}\right)}{4+\sqrt{4-x^2}}\)
\(\Rightarrow A=\sqrt{\left(2+x\right)^{^{ }3}}-\sqrt{\left(2-x\right)^3}=\left(\sqrt{2+x}-\sqrt{2-x}\right)\left(4+\sqrt{4-x^2}\right)\)
\(\Rightarrow A=\dfrac{\sqrt{4+2\sqrt{4-x^2}}\left(\sqrt{2+x}-\sqrt{2-x}\right)\left(4+\sqrt{4-x^2}\right)}{\sqrt{2}\left(4+\sqrt{4-x^2}\right)}\)
\(\Rightarrow A=\dfrac{\left(\sqrt{2+x}+\sqrt{2-x}\right)\left(\sqrt{2+x}-\sqrt{2-x}\right)}{\sqrt{2}}=2\sqrt{2}\)
1/ a/ \(\sqrt{\left(6+2\sqrt{5}\right)^3}-\sqrt{\left(6-2\sqrt{5}\right)^3}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^6}-\sqrt{\left(\sqrt{5}-1\right)^6}\)
\(=\left(\sqrt{5}+1\right)^3-\left(\sqrt{5}-1\right)^3\)
\(=32\)
b/ \(\sqrt{\left(3-2\sqrt{2}\right)\left(4-2\sqrt{3}\right)}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{2}-1\right)\left(\sqrt{3}-1\right)\)
\(=\sqrt{6}-\sqrt{2}-\sqrt{3}+1\)
Câu 3/ \(A=\sqrt{2+\sqrt{2+\sqrt{2+...+\sqrt{2+\sqrt{2}}}}}\)
\(< \sqrt{2+\sqrt{2+\sqrt{2+...+\sqrt{2+\sqrt{4}}}}}=2\)
Ta lại có:
\(A=\sqrt{2+\sqrt{2+\sqrt{2+...+\sqrt{2+\sqrt{2}}}}}>\sqrt{2}>1\)
\(\Rightarrow1< A< 2\)
Vậy \(A\notin N\)
\(\sqrt{4\left(a-3\right)^2}\)
\(=\sqrt{2^2\left(a-3\right)^2}\)
\(=2\left(a-3\right)\)
\(=2a-6\)
\(\sqrt{4\left(a-3\right)^2}=\sqrt{\left[2\left(a-3\right)\right]^2}=2\left(a-3\right)\)3)