a) A = \(13-2\sqrt{42}=\left(\sqrt{7}-\sqrt{6}\right)^2\)

<=> \(\sqrt{A}=\sqrt{7}-\sqrt{6}\)

b) \(A=46+6\sqrt{5}=\left(\sqrt{45}+1\right)^2\)

<=> \(\sqrt{A}=\sqrt{45}+1\)

c) \(A=12-3\sqrt{15}=\dfrac{1}{2}\left(24-6\sqrt{15}\right)=\dfrac{1}{2}\left(\sqrt{15}-3\right)^2\)

<=> \(\sqrt{A}=\dfrac{1}{\sqrt{2}}\left(\sqrt{15}-3\right)\)

\(\sqrt{13-2\sqrt{42}}=\sqrt{6-2\sqrt{6}.\sqrt{7}+7}=\sqrt{\left(\sqrt{6}-\sqrt{7}\right)^2}=\left|\sqrt{6}-\sqrt{7}\right|=\sqrt{7}-\sqrt{6}\)

\(\sqrt{46+6\sqrt{5}}=\sqrt{45+6\sqrt{5}+1}=\sqrt{3^2.5+6\sqrt{5}+1}=\sqrt{3^2.5+2.3.\sqrt{5}+1^2}=\sqrt{\left(3.\sqrt{5}+1\right)^2}=3\sqrt{5}+1\)

\(\sqrt{12-3\sqrt{15}}=\sqrt{3}\sqrt{4-\sqrt{15}}=\sqrt{\frac{3}{2}}.\sqrt{8-2\sqrt{15}}=\sqrt{\frac{3}{2}}.\sqrt{3-2\sqrt{15}+5}=\sqrt{\frac{3}{2}}.\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{\frac{3}{2}}.\left(\sqrt{5}-\sqrt{3}\right)\)

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

\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\sqrt{\frac{1}{2}}\left(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\right)=\sqrt{\frac{1}{2}}\left(\sqrt{1+2\sqrt{5}+5}-\sqrt{1-2\sqrt{5}+5}\right)=\sqrt{\frac{1}{2}}\left(\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\right)=\sqrt{\frac{1}{2}}\left(1+\sqrt{5}-\sqrt{5}+1\right)=\sqrt{\frac{1}{2}}.2=\sqrt{\frac{4}{2}}=\sqrt{2}\)

+6

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

=>\(\sqrt{A}=\sqrt{7}-\sqrt{6}\)

+2.\(3\sqrt{5}\)+1

=(\(3\sqrt{5}+1\))²

=>\(\sqrt{A}=3\sqrt{5}+1\)

Bài 2 : 

a,\(\sqrt{24}+\sqrt{45}< \sqrt{25}+\sqrt{49}=5+7=12=>\sqrt{24}+\sqrt{45}< 12\)

b. \(\sqrt{37}-\sqrt{15}>\sqrt{36}-\sqrt{16}=6-4=2=>\sqrt{37}-\sqrt{15}>2\)

c, \(\sqrt{15}.\sqrt{17}>\sqrt{15}.\sqrt{16}>\sqrt{16}=>\sqrt{15}.\sqrt{17}>\sqrt{16}\)

\(A=\sqrt{13+4\sqrt{10}}=\sqrt{13+2\sqrt{40}}=\sqrt{8+2.\sqrt{5}.\sqrt{8}+5}=\sqrt{\left(\sqrt{8}+\sqrt{5}\right)^2}=\sqrt{8}+\sqrt{5}\)

\(B=\sqrt{46-6\sqrt{5}}=\sqrt{46-2\sqrt{45}}=\sqrt{\left(\sqrt{45}-1\right)^2}=\sqrt{45}-1=3\sqrt{5}-1\)

\(C=\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{1}{\sqrt{3}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-\sqrt{7}}\)

\(C=-\sqrt{3}-\sqrt{2}+\dfrac{\sqrt{5}+\sqrt{3}}{2}-\dfrac{\sqrt{7}+\sqrt{5}}{2}\)

\(C=-\sqrt{3}-\sqrt{2}+\dfrac{\sqrt{3}-\sqrt{7}}{2}\)

\(C=\dfrac{-2\sqrt{3}-2\sqrt{2}+\sqrt{3}-\sqrt{7}}{2}=\dfrac{-\sqrt{3}-2\sqrt{2}-\sqrt{7}}{2}\)

Mk làm đc có 3 câu thôi.

D = (4\(\sqrt{10}\) - 4\(\sqrt{6}\) + 5\(\sqrt{6}\) - 3\(\sqrt{10}\) )\(\sqrt{4-\sqrt{15}}\)

D = (\(\sqrt{10}\) + \(\sqrt{6}\) )\(\sqrt{4-\sqrt{15}}\)

D = \(\sqrt{\left(4-\sqrt{15}\right)10}\) + \(\sqrt{\left(4-\sqrt{15}\right)6}\)

D = \(\sqrt{40-10\sqrt{15}}\) + \(\sqrt{24-6\sqrt{15}}\)

D = \(\sqrt{\left(\sqrt{15}\right)^2-2.5.\sqrt{5}+5^2}\) + \(\sqrt{\left(\sqrt{15}\right)^2-2.3.\sqrt{15}+3^2}\)

D = \(\sqrt{\left(\sqrt{15}-5\right)^2}\) + \(\sqrt{\left(\sqrt{15}-3\right)^2}\)

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

F=\(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)

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

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

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

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

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

=1

* \(\sqrt{2}\)A = \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}+\sqrt{14}=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{14}=\sqrt{7}-1-\left(\sqrt{7}+1\right)+\sqrt{14}=\sqrt{14}-2\)

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

* B là 6,5 hay 6*5 vậy bạn

nếu 6,5 thì : B cũng nhân \(\sqrt{2}\) biểu thức trở thành

\(\sqrt{2}B=\sqrt{13+2\sqrt{12}}+\sqrt{13-2\sqrt{12}}+4\sqrt{3}=\sqrt{\left(1+\sqrt{12}\right)^2}+\sqrt{\left(\sqrt{12}-1\right)^2}+4\sqrt{3}=1+\sqrt{12}+\sqrt{12}-1+4\sqrt{3}=4\sqrt{3}+4\sqrt{3}=8\sqrt{3}\)

=> B = \(\dfrac{8\sqrt{3}}{\sqrt{2}}=4\sqrt{6}\)

nếu 6*5 thì : bạn tách hai căn đầu thành một biểu thức rồi bình phương lên rồi giải , sau đó trục căn , biểu thức luôn dương nhé , mấy bài này nếu không thể tách thì làm cách này cũng được

* C thì mik chỉ bít pt được nhiu đây thôi , bạn thông cảm nhé\(\sqrt{29-6\sqrt{20}}=\sqrt{\left(\sqrt{20}-3\right)^2}=\sqrt{20}+3=2\sqrt{5}-3\)

* D = \(\sqrt{13-2\cdot2\sqrt{2}\cdot\sqrt{5}}-\sqrt{53+2\cdot2\sqrt{2}\cdot3\sqrt{5}}=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(2\sqrt{2}+3\sqrt{5}\right)^2}=2\sqrt{2}-\sqrt{5}-2\sqrt{2}-3\sqrt{5}=-4\sqrt{5}\)

Câu C có sai đề ko? Tui sửa đây!

Ta có: \(C=\sqrt{46+6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

=> \(C=\sqrt{45+2.3\sqrt{5}+1}-\sqrt{20-2.3.2\sqrt{5}+9}\)

=> \(C=\sqrt{\left(3\sqrt{5}+1\right)^2}-\sqrt{\left(2\sqrt{5}-3\right)^2}\)

=> \(C=\left|3\sqrt{5}+1\right|-\left|2\sqrt{5}-3\right|\)

=> \(C=3\sqrt{5}+1-2\sqrt{5}+3=4+\sqrt{5}\)

\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3\sqrt{3}-2\sqrt{2}\right)^2}\)

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

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

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

\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\sqrt{\left(5-2\sqrt{6}\right)^2}+\sqrt{\left(5+2\sqrt{6}\right)^2}\)

\(=5-2\sqrt{6}+5+2\sqrt{6}=10\)

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

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

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

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

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

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

=6

c: Ta có: \(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)

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

=10

d: Ta có: \(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}\)

\(=\sqrt{13-4\sqrt{10}}+\sqrt{53+4\sqrt{90}}\)

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

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

Bài 1:

a)

\(\sqrt{13-2\sqrt{42}}=\sqrt{6+7-2\sqrt{6.7}}=\sqrt{(\sqrt{7}-\sqrt{6})^2}=|\sqrt{7}-\sqrt{6}|=\sqrt{7}-\sqrt{6}\)

b)

\(\sqrt{46+6\sqrt{5}}=\sqrt{46+2\sqrt{45}}=\sqrt{45+1+2\sqrt{45.1}}=\sqrt{(\sqrt{45}+1)^2}=\sqrt{45}+1\)

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

c)

\(\sqrt{12-3\sqrt{15}}=\sqrt{\frac{24-6\sqrt{15}}{2}}=\sqrt{\frac{24-2\sqrt{135}}{2}}=\sqrt{\frac{15+9-2\sqrt{15.9}}{2}}\)

\(=\sqrt{\frac{(\sqrt{15}-\sqrt{9})^2}{2}}=\frac{\sqrt{15}-\sqrt{9}}{\sqrt{2}}=\frac{\sqrt{15}-3}{\sqrt{2}}\)

d)

\(\sqrt{11+\sqrt{96}}=\sqrt{11+2\sqrt{24}}=\sqrt{8+3+2\sqrt{8.3}}\)

\(=\sqrt{(\sqrt{8}+\sqrt{3})^2}=\sqrt{8}+\sqrt{3}\)

Bài 1:

a)

\(\sqrt{13-2\sqrt{42}}=\sqrt{6+7-2\sqrt{6.7}}=\sqrt{(\sqrt{7}-\sqrt{6})^2}=|\sqrt{7}-\sqrt{6}|=\sqrt{7}-\sqrt{6}\)

b)

\(\sqrt{46+6\sqrt{5}}=\sqrt{46+2\sqrt{45}}=\sqrt{45+1+2\sqrt{45.1}}=\sqrt{(\sqrt{45}+1)^2}=\sqrt{45}+1\)

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

c)

\(\sqrt{12-3\sqrt{15}}=\sqrt{\frac{24-6\sqrt{15}}{2}}=\sqrt{\frac{24-2\sqrt{135}}{2}}=\sqrt{\frac{15+9-2\sqrt{15.9}}{2}}\)

\(=\sqrt{\frac{(\sqrt{15}-\sqrt{9})^2}{2}}=\frac{\sqrt{15}-\sqrt{9}}{\sqrt{2}}=\frac{\sqrt{15}-3}{\sqrt{2}}\)

d)

\(\sqrt{11+\sqrt{96}}=\sqrt{11+2\sqrt{24}}=\sqrt{8+3+2\sqrt{8.3}}\)

\(=\sqrt{(\sqrt{8}+\sqrt{3})^2}=\sqrt{8}+\sqrt{3}\)

Bài 2:

a)

\(A=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3+1+2\sqrt{3}}}}\)

\(=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{(\sqrt{3}+1)^2}}}=\sqrt{6+2\sqrt{2}.\sqrt{2-\sqrt{3}}}\)

\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{(\sqrt{3}-1)^2}}=\sqrt{6+2(\sqrt{3}-1)}\)

\(=\sqrt{4+2\sqrt{3}}=\sqrt{3+1+2\sqrt{3.1}}=\sqrt{(\sqrt{3}+1)^2}=\sqrt{3}+1\)

b)

\(B=\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}=\sqrt{5}-\sqrt{3-\sqrt{20+9-2\sqrt{20.9}}}\)

\(=\sqrt{5}-\sqrt{3-\sqrt{(\sqrt{20}-\sqrt{9})^2}}=\sqrt{5}-\sqrt{3-(\sqrt{20}-\sqrt{9})}\)

\(=\sqrt{5}-\sqrt{6-2\sqrt{5}}=\sqrt{5}-\sqrt{5+1-2\sqrt{5}}=\sqrt{5}-\sqrt{(\sqrt{5}-1)^2}\)

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