\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\sqrt{25-2\cdot5\cdot2\sqrt{6}+24}+\sqrt{25-2\cdot5\cdot2\sqrt{6}+24}=\sqrt{\left(5+2\sqrt{6}\right)^2}+\sqrt{\left(5-2\sqrt{6}\right)^2}=5+2\sqrt{6}+5-2\sqrt{6}=10\) ---

\(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}=\sqrt{8-2\sqrt{5}\cdot\sqrt{8}+5}+\sqrt{45+2\cdot3\sqrt{5}\cdot\sqrt{8}+8}=\sqrt{\left(\sqrt{8}-\sqrt{5}\right)^2}+\sqrt{\left(3\sqrt{5}+\sqrt{8}\right)^2}=\sqrt{8}-\sqrt{5}+3\sqrt{5}+\sqrt{8}=2\sqrt{8}+2\sqrt{5}\)

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\(\sqrt{11-6\sqrt{2}}+\sqrt{3-2\sqrt{2}}=\sqrt{9-2\cdot3\cdot\sqrt{2}+2}+\sqrt{2-2\sqrt{2}+1}=\sqrt{\left(3-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}-1\right)^2}=3-\sqrt{2}+\sqrt{2}-1=2\)

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\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}=\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{27-2\cdot\sqrt{27}\cdot\sqrt{8}+8}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3\sqrt{3}-2\sqrt{2}\right)^2}=3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)

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\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{9-2\cdot3\cdot2\sqrt{2}+8}+\sqrt{9+2\cdot2\cdot2\sqrt{2}+8}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}=3-2\sqrt{2}+3+2\sqrt{2}=6\)

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\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)

\(=\sqrt{25-2\times5\sqrt{24}+24}+\sqrt{25+2\times5\sqrt{24}+24}\)

\(=\sqrt{\left(5-\sqrt{24}\right)^2}+\sqrt{\left(5+\sqrt{24}\right)^2}\)

\(=5-\sqrt{24}+5+\sqrt{24}\)

\(=10\)

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

\(=\sqrt{9-2\times3\sqrt{2}+2}+\sqrt{2-2\times1\sqrt{2}+1}\)

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

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

\(=2\)

a, \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)

= \(\sqrt{3^2-2.3.\sqrt{6}+\left(\sqrt{6}\right)^2}+\sqrt{6^2-2.6.\sqrt{6}+\left(\sqrt{6}\right)^2}\)

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

= \(\left|3-\sqrt{6}\right|+\left|6-\sqrt{6}\right|\)

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

= \(9-2\sqrt{6}\)

b. Đặt B = \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)

Nhận xét : B > 0 , bình phương hai vế ta được :

\(B^2=\left(\sqrt{17-3\sqrt{32}}\right)^2+\left(\sqrt{17+3\sqrt{32}}\right)^2\)

\(B^2=17-3\sqrt{32}+17+3\sqrt{32}+2\sqrt{\left(17-3\sqrt{32}\right)\left(17+3\sqrt{32}\right)}\)

\(B^2=34+2\sqrt{17^2-\left(3\sqrt{32}\right)^2}\)

\(B^2=34+2\sqrt{289-288}\)

\(B^2=34+2=36\)

=> \(B=\pm\sqrt{36}\) mà B > 0 nên \(B=\sqrt{36}=6\)

c, Đặt C = \(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)

Nhận xét : C > 0 , bình phương hai vế ta đươc :

\(C^2=\left(\sqrt{49-5\sqrt{96}}\right)^2+\left(\sqrt{49+5\sqrt{96}}\right)^2\)

\(C^2=49-5\sqrt{96}+49+5\sqrt{96}+2\sqrt{\left(49-5\sqrt{96}\right)\left(49+5\sqrt{96}\right)}\)

\(C^2=98+2\sqrt{49^2-\left(5\sqrt{96}\right)^2}\)

\(C^2=98+2\sqrt{2401-2400}\)

\(C^2=98+2=100\)

=> \(C=\pm\sqrt{100}\) mà C > 0 nên \(C=\sqrt{100}=10\)

a) Ta có: \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)

\(=\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{27-2\cdot3\sqrt{3}\cdot2\sqrt{2}+8}\)

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

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

\(=3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)(Vì \(\left\{{}\begin{matrix}3>\sqrt{6}\\3\sqrt{3}>2\sqrt{2}\end{matrix}\right.\))

b) Ta có: \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)

\(=\frac{\sqrt{34-6\sqrt{32}}+\sqrt{34+6\sqrt{32}}}{\sqrt{2}}\)

\(=\frac{\sqrt{18-2\cdot3\sqrt{2}\cdot4+16}+\sqrt{18+2\cdot3\sqrt{2}\cdot4+16}}{\sqrt{2}}\)

\(=\frac{\sqrt{\left(3\sqrt{2}-4\right)^2}+\sqrt{\left(3\sqrt{2}+4\right)^2}}{\sqrt{2}}\)

\(=\frac{\left|3\sqrt{2}-4\right|+\left|3\sqrt{2}+4\right|}{\sqrt{2}}\)

\(=\frac{3\sqrt{2}-4+3\sqrt{2}+4}{\sqrt{2}}\)(Vì \(3\sqrt{2}>4>0\))

\(=\frac{6\sqrt{2}}{\sqrt{2}}=6\)

\(A=\left(2-\sqrt{3}\right)\sqrt{4+2.2.\sqrt{3}+3}=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=1\)

các câu còn lại làm tương tự nhé bạn !

câu đầu có \(3-12\sqrt{6}< 0\) nên không căn được nên đề bạn sai

\(\sqrt{31-8\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)

\(=\sqrt{4^2-2.4.\sqrt{15}+\left(\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}\right)^2-2.\sqrt{15}.3+3^2}\)

\(=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}=\left|4-\sqrt{15}\right|+\left|\sqrt{15}-3\right|\)

\(=4-\sqrt{15}+\sqrt{15}-3=1\)

\(\sqrt{49-5\sqrt{96}}-\sqrt{49+5\sqrt{96}}=\sqrt{49-20\sqrt{6}}-\sqrt{49+20\sqrt{6}}\)

\(=\sqrt{5^2-2.5.2\sqrt{6}+\left(2\sqrt{6}\right)^2}-\sqrt{5^2+2.5.4\sqrt{6}+\left(2\sqrt{6}\right)^2}\)

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

\(=5-2\sqrt{6}-5-2\sqrt{6}=-4\sqrt{6}\)

\(\sqrt{31-8\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)

\(=4-\sqrt{15}+\sqrt{15}-3\)

=1

32, \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

=\(\sqrt{9-2.3.\sqrt{6}+6}+\sqrt{33-2.3.2\sqrt{6}}\)

=\(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{24-2.3.2\sqrt{6}+9}\)

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

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

33, \(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}=\left|\sqrt{5}-1\right|+\sqrt{5}+1=\sqrt{5}-1+\sqrt{5}+1=2\sqrt{5}\)

34, \(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}\)

=\(\sqrt{8-2.\sqrt{3}.\sqrt{5}}-\sqrt{23-2.2.\sqrt{5}.\sqrt{3}}\)

=\(\sqrt{5-2\sqrt{3}.\sqrt{5}+3}-\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}.\sqrt{3}+3}\)

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

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

35,\(\sqrt{31-8\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)

=\(\sqrt{16-2.4.\sqrt{15}+15}+\sqrt{15-2.3.\sqrt{15}+9}\)

=\(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)

=\(\left|4-\sqrt{15}\right|+\left|\sqrt{15}-3\right|\)

=\(4-\sqrt{15}+\sqrt{15}-3\)

=1

36, \(\sqrt{49-5\sqrt{96}}-\sqrt{49+5\sqrt{96}}\)

=\(\sqrt{49-2.5.\sqrt{24}}-\sqrt{49+2.5\sqrt{24}}=\sqrt{25-2.5.\sqrt{24}+24}-\sqrt{25+2.5.\sqrt{24}+24}=\sqrt{\left(5-\sqrt{24}\right)^2}-\sqrt{\left(5+\sqrt{24}\right)^2}\)

=\(\left|5-\sqrt{24}\right|-\left|5+\sqrt{24}\right|=5-\sqrt{24}-5-\sqrt{24}=-2\sqrt{24}\)

37, \(\sqrt{3+2\sqrt{2}}+\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)

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

f, \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}+\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=\sqrt{\sqrt{5}+\sqrt{3-2\sqrt{5}+3}}=\sqrt{\sqrt{5}+\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}+\sqrt{5}-1}=\sqrt{2\sqrt{5}-1}\)

mik sửa lại câu f , tí nhé :

f , \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

a,\(=\sqrt{9-2.3.2\sqrt{2}+8}-\sqrt{9+2.3.2\sqrt{2}+8}\)

\(=\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(3+2\sqrt{2}\right)^2}\) \(=3-2\sqrt{2}-3-2\sqrt{2}=-4\sqrt{2}\)