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

\(\sqrt{29-4\sqrt{7}}=\sqrt{\left(2\sqrt{7}\right)^2-2.2\sqrt{7}.1+1^2}=\sqrt{\left(2\sqrt{7}-1\right)^2}=\left|2\sqrt{7}-1\right|\)

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

\(\sqrt{19+6\sqrt{2}}=\sqrt{\left(3\sqrt{2}\right)^2+2.3\sqrt{2}.1+1^2}=\sqrt{\left(3\sqrt{2}+1\right)^2}=\left|3\sqrt{2}+1\right|\)

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

\(\sqrt{28-6\sqrt{3}}=\sqrt{\left(3\sqrt{3}\right)^2-2.3\sqrt{3}.1+1^2}=\sqrt{\left(3\sqrt{3}-1\right)^2}=\left|3\sqrt{3}-1\right|\)

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

\(\sqrt{46-6\sqrt{5}}=\sqrt{\left(3\sqrt{5}\right)^2-2.3\sqrt{5}.1+1^2}=\sqrt{\left(3\sqrt{5}-1\right)^2}=\left|3\sqrt{5}-1\right|\)

\(=3\sqrt{5}-1\)

\(\sqrt{49+8\sqrt{3}}=\sqrt{\left(4\sqrt{3}\right)^2+2.4\sqrt{3}.1+1^2}=\sqrt{\left(4\sqrt{3}+1\right)^2}=\left|4\sqrt{3}+1\right|\)

\(=4\sqrt{3}+1\)

\(\sqrt{32-8\sqrt{7}}=\sqrt{\left(2\sqrt{7}\right)^2-2.2\sqrt{7}.2+2^2}=\sqrt{\left(2\sqrt{7}-2\right)^2}=\left|2\sqrt{7}-2\right|\)

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

\(\sqrt{29-4\sqrt{7}}=2\sqrt{7}-1\)

\(\sqrt{19+6\sqrt{2}}=3\sqrt{2}+1\)

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

\(\sqrt{46-6\sqrt{5}}=3\sqrt{5}-1\)

\(\sqrt{49+8\sqrt{3}}=4\sqrt{3}+1\)

\(\sqrt{32-8\sqrt{7}}=2\sqrt{7}-2\)

a: Ta có: \(\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\)

\(=\sqrt{5}+\sqrt{3}-\sqrt{5}-1\)

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

b: Ta có: \(\sqrt{17-2\sqrt{72}}+\sqrt{19+2\sqrt{18}}\)

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

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

c: Ta có: \(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\)

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

\(=4\sqrt{2}-1\)

a)

\(\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{5+2\sqrt{5}\cdot\sqrt{3}+3}-\sqrt{5+2\sqrt{5}\cdot\sqrt{1}+1}\\ =\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{1}\right)^2}\\ =\sqrt{5}+\sqrt{3}-\sqrt{5}-\sqrt{1}\\ =\sqrt{3}-\sqrt{1}\)

b)

\(\sqrt{17-2\sqrt{72}}+\sqrt{19+2\sqrt{18}}\\ =\sqrt{9-2\sqrt{9}\cdot\sqrt{8}+8}+\sqrt{18+2\sqrt{18}\cdot\sqrt{1}+1}\\ =\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}+1\right)^2}\\ =3-2\sqrt{2}+3\sqrt{2}+1\\ =4+\sqrt{2}\)

c)

\(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\\ =\sqrt{8-2\sqrt{8}\cdot\sqrt{4}+4}+\sqrt{8+2\sqrt{8}\cdot\sqrt{1}+1}\\ =\sqrt{\left(2\sqrt{2}-2\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}\\ =2\sqrt{2}-2+2\sqrt{2}+1\\ =4\sqrt{2}-1\)

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

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

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

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

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

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

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

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

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

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

1)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\sqrt{11}-\sqrt{3}\)
2)
\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}=\sqrt{7}-\sqrt{5}\)
3)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{11}-\sqrt{5}\right)}=\sqrt{11}-\sqrt{5}\)
4)
\(=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
5)
\(=\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\)

 

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

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

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

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

6) \(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\)

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

a) Ta có: \(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)

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

=-4

b) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)

\(=\sqrt{3}-1-2+\sqrt{3}+4+\sqrt{3}\)

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

c) Ta có: \(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)

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

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

d) Ta có: \(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2\right)^6}\)

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

=10

a) \(\sqrt{14-6\sqrt{5}}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)

b, c) tương tự câu a.

d) \(\left(3-\sqrt{2}\right)\sqrt{11+6\sqrt{2}}\)

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

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

\(=9-2\)

\(=7\)

e) \(\sqrt{11-6\sqrt{2}+\sqrt{3-2\sqrt{2}}}\)

\(=\sqrt{11-6\sqrt{2}+\sqrt{\left(1-\sqrt{2}\right)^2}}\)

\(=\sqrt{11-6\sqrt{2}+\sqrt{2}-1}\)

\(=\sqrt{10-5\sqrt{2}}\)

Ok !! chi tiết =))

\(\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}-\sqrt{4+2\sqrt{3}}\)

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

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

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

\(=\sqrt{2}\)