\(\sqrt{12-6\sqrt{3}}=\sqrt{9-6\sqrt{3}+3}=\sqrt{3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(3-\sqrt{3}\right)^2}\)

\(=\left|3-\sqrt{3}\right|=3-\sqrt{3}\)

\(\sqrt{19+8\sqrt{3}}=\sqrt{16+8\sqrt{3}+3}=\sqrt{4^2+2.4.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(4+\sqrt{3}\right)^2}\)

\(=\left|4+\sqrt{3}\right|=4+\sqrt{3}\)

\(\sqrt{14-6\sqrt{5}}=\sqrt{9-6\sqrt{5}+5}=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}\)

\(=\left|3-\sqrt{5}\right|=3-\sqrt{5}\)

\(\sqrt{12-6\sqrt{3}}=\sqrt{3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(3-\sqrt{3}\right)^2}=\left|3-\sqrt{3}\right|=3-\sqrt{3}\)

\(\sqrt{19+8\sqrt{3}}=\sqrt{4^2+2.4.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(4+\sqrt{3}\right)^2}=\left|4+\sqrt{3}\right|=4+\sqrt{3}\)

\(\sqrt{14-6\sqrt{5}}=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=\left|3-\sqrt{5}\right|=3-\sqrt{5}\)

\(\sqrt{12-6\sqrt{3}}=3-\sqrt{3}\)

\(\sqrt{19+8\sqrt{3}}=4+\sqrt{3}\)

\(\sqrt{14-6\sqrt{5}}=3-\sqrt{5}\)

1) \(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)=\left(\sqrt{19}\right)^2-3^2=19-9=10\)

2) \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\sqrt{\dfrac{8+2\sqrt{7}}{2}}-\sqrt{\dfrac{8-2\sqrt{7}}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{7}\right)^2+2.\sqrt{7}.1+1^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}\right)^2-2.\sqrt{7}.1+1^2}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{7}+1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}-1\right)^2}{2}}=\dfrac{\left|\sqrt{7}+1\right|}{\sqrt{2}}-\dfrac{\left|\sqrt{7}-1\right|}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}+1}{\sqrt{2}}-\dfrac{\sqrt{7}-1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)

3) \(\sqrt{8+\sqrt{60}}+\sqrt{45}-\sqrt{12}=\sqrt{8+\sqrt{4.15}}+\sqrt{9.5}-\sqrt{4.3}\)

\(=\sqrt{8+2\sqrt{15}}+3\sqrt{5}-2\sqrt{3}\)

\(=\sqrt{\left(\sqrt{5}\right)^2+2.\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}+3\sqrt{5}-2\sqrt{3}\)

\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+3\sqrt{5}-2\sqrt{3}=\left|\sqrt{5}+\sqrt{3}\right|+3\sqrt{5}-2\sqrt{3}\)

\(\sqrt{5}+\sqrt{3}+3\sqrt{5}-2\sqrt{3}=4\sqrt{5}-\sqrt{3}\)

4) \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)

\(=\sqrt{\left(\sqrt{5}\right)^2-2.2.\sqrt{5}+2^2}-\sqrt{\left(\sqrt{5}\right)^2+2.2.\sqrt{5}+2^2}\)

\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}=\left|\sqrt{5}-2\right|-\left|\sqrt{5}+2\right|\)

\(=\sqrt{5}-2-\sqrt{5}-2=-4\)

1) \(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)=19-9=10\)

4) \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}=\sqrt{5}-2-\sqrt{5}-2=-4\)

=\(\sqrt{3+2\sqrt{3}+1}\)+\(\sqrt{3-2\sqrt{3}+1}\)

=\(\sqrt{\left(\sqrt{3}+1\right)^2}\)+\(\sqrt{\left(\sqrt{3}-1\right)^2}\)

=\(\sqrt{3}+1+\sqrt{3}-1\)

=\(2\sqrt{3}\)

k mk nha

\(=\left(-1\right)\sqrt{\left(\sqrt{3}+\sqrt{1}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}\)

\(=\left(-1\right)\cdot\left(\sqrt{3}+1\right)+\left(\sqrt{3}-1\right)\)

\(=\left(-\sqrt{3}-1\right)+\left(\sqrt{3}-1\right)\)

\(=-2\)

\(\sqrt{7-4\sqrt{3}}=\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)

\(\sqrt{9+4\sqrt{5}}=\sqrt{2^2+2.2.\sqrt{5}+\left(\sqrt{5}\right)^2}+\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2+\sqrt{5}\right|=2+\sqrt{5}\)

\(\sqrt{11-4\sqrt{7}}=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}.2+2^2}=\sqrt{\left(\sqrt{7}-2\right)^2}=\left|\sqrt{7}-2\right|=\sqrt{7}-2\)

\(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}\)

\(\sqrt{9+4\sqrt{5}}=\sqrt{5}+2\)

\(\sqrt{11-4\sqrt{7}}=\sqrt{7}-2\)

Lần sau bạn chú ý viết đầy đủ đề.

1.

\(\sqrt{9+4\sqrt{5}-\sqrt{9-4\sqrt{5}}}=\sqrt{9+4\sqrt{5}-\sqrt{5-2\sqrt{4.5}+4}}\)

\(=\sqrt{9+4\sqrt{5}-\sqrt{(\sqrt{5}-\sqrt{4})^2}}=\sqrt{9+4\sqrt{5}-(\sqrt{5}-\sqrt{4})}\)

\(=\sqrt{9+4\sqrt{5}-\sqrt{5}+2}=\sqrt{11+3\sqrt{5}}\)

2.

\(\sqrt{8-2\sqrt{7}-\sqrt{8+2\sqrt{7}}}=\sqrt{8-2\sqrt{7}-\sqrt{7+2\sqrt{7}+1}}\)

\(=\sqrt{8-2\sqrt{7}-\sqrt{(\sqrt{7}+1)^2}}\)

\(=\sqrt{8-2\sqrt{7}-\sqrt{7}-1}=\sqrt{7-3\sqrt{7}}\)

Phạm Mạnh Kiên: sửa lại theo ý bạn thì làm như sau:

1.

\(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}=\sqrt{5+2\sqrt{5}.\sqrt{4}+4}-\sqrt{5-2\sqrt{5}.\sqrt{4}+4}\)

\(=\sqrt{(\sqrt{5}+\sqrt{4})^2}-\sqrt{(\sqrt{5}-\sqrt{4})^2}=|\sqrt{5}+2|-|\sqrt{5}-2|\)

\(=\sqrt{5}+2-(\sqrt{5}-2)=4\)

2.

\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)

\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}=|\sqrt{7}-1|-|\sqrt{7}+1|\)

\(=-2\)

\(\sqrt{13+\sqrt{48}}=\sqrt{13+\sqrt{4.12}}=\sqrt{13+2\sqrt{12}}=\sqrt{\left(\sqrt{12}+1\right)^2}\)

\(=\sqrt{12}+1=2\sqrt{3}+1\)

\(\Rightarrow\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=\sqrt{3}-1\)

\(\Rightarrow\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{3+\sqrt{3}-1}=\sqrt{2+\sqrt{3}}\)

\(\Rightarrow\sqrt{\dfrac{4+2\sqrt{3}}{2}}=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}\)

\(\Rightarrow2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}==2.\dfrac{\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}+\sqrt{2}\)

2) biến đổi khúc sau như câu 1:

\(\Rightarrow\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}\)

\(=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

4) Ta có: \(\sqrt{30-2\sqrt{16+6\sqrt{11+4\sqrt{4-2\sqrt{3}}}}}\)

\(=\sqrt{30-2\sqrt{16+6\sqrt{11+4\left(\sqrt{3}-1\right)}}}\)

\(=\sqrt{30-2\sqrt{16+6\sqrt{7+4\sqrt{3}}}}\)

\(=\sqrt{30-2\sqrt{16+6\left(2+\sqrt{3}\right)}}\)

\(=\sqrt{30-2\sqrt{28+6\sqrt{3}}}\)

\(=\sqrt{30-2\left(3\sqrt{3}+1\right)}\)

\(=\sqrt{28-6\sqrt{3}}=3\sqrt{3}-1\)

5) Ta có: \(\dfrac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)

\(=\dfrac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{\sqrt{75}-5\sqrt{2}}\)

\(=\dfrac{5\left(\sqrt{3}-\sqrt{2}\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}=1\)