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

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

+6

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

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

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

=(\(3\sqrt{5}+1\))²

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

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

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

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

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

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

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

\(46+6\sqrt{5}=\left(5+2\cdot\sqrt{5}\cdot3+9\right)+32=\left(\sqrt{5}+3\right)^2+32\)(ko rút đc)

\(\sqrt{3-\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\\ =\sqrt{3-\sqrt{5}}\cdot\sqrt{2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\\ =\sqrt{6-2\sqrt{5}}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\\ =\sqrt{5-2\sqrt{5}+1}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\\ =\sqrt{\left(\sqrt{5}-1\right)^2}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\\ =\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\left(3+\sqrt{5}\right)\\ =4\left(3+\sqrt{5}\right)\)

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

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

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

Dễ dàng nhận ra

\(\sqrt{\sqrt{7}-\sqrt{3}}< \sqrt{\sqrt{7}+\sqrt{3}}\Rightarrow\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}< 0\)

Đặt \(x=\frac{\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}}{\sqrt{\sqrt{7}-2}}< 0\)

\(\Rightarrow x^2=\frac{\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}-2\sqrt{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}}{\sqrt{7}-2}\)

\(\Rightarrow x^2=\frac{2\sqrt{7}-2\sqrt{4}}{\sqrt{7}-2}=\frac{2\sqrt{7}-4}{\sqrt{7}-2}=\frac{2\left(\sqrt{7}-2\right)}{\sqrt{7}-2}=2\)

\(\Rightarrow x=-\sqrt{2}\) (do \(x< 0\))

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

(đến đây bó tay :<)

\(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\\ =\sqrt{5}-\sqrt{3-\sqrt{20-2\cdot2\sqrt{5}\cdot3+9}}\\ =\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot3+3^2}}\\ =\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}\\ =\sqrt{5}-\sqrt{3-2\sqrt{5}+3}=\\ \sqrt{5}-\sqrt{6-2\sqrt{5}}\\ =\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}\\ =\sqrt{5}-\sqrt{5}+1=1\)

(cái này lỡ tính rùi :3)

Câu cuối không biết, sorry nha

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

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

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

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

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

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

* \(\sqrt{2}\)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: \(C=\sqrt{46+6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

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

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

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

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

Bài 2:

a) \(2\sqrt{125}+\dfrac{3}{2}\sqrt{80}-\sqrt{180}-\dfrac{2}{7}\sqrt{245}\)

\(=2\sqrt{5^2\cdot5}+\dfrac{3}{2}\sqrt{4^2\cdot5}-\sqrt{6^2\cdot5}-\dfrac{2}{7}\sqrt{7^2\cdot5}\)

\(=10\sqrt{5}+\dfrac{3\cdot4}{2}\sqrt{5}-6\sqrt{5}-\dfrac{2\cdot7}{7}\sqrt{5}\)

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

\(=8\sqrt{5}\)

b) \(\sqrt{11-4\sqrt{7}}-\sqrt{16+6\sqrt{7}}\)

\(=\sqrt{\left(\sqrt{7}\right)^2-2\cdot2\cdot\sqrt{7}+2^2}-\sqrt{\left(\sqrt{7}\right)^2+2\cdot3\cdot\sqrt{7}+3^2}\)

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

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

\(=-5\)

\(2a,\\ 2\sqrt{125}+\dfrac{3}{2}.\sqrt{80}-\sqrt{180}-\dfrac{2}{7}\sqrt{245}\\ =2\sqrt{5^2.5}+\dfrac{3}{2}.\sqrt{4^2.5}-\sqrt{6^2.5}-\dfrac{2}{7}.\sqrt{7^2.5}\\ =2.5.\sqrt{5}+\dfrac{3}{2}.4.\sqrt{5}-6\sqrt{5}-\dfrac{2}{7}.7\sqrt{5}\\ =10\sqrt{5}+6\sqrt{5}-6\sqrt{5}-2\sqrt{5}=8\sqrt{5}\)

3: 

a: =>|x-1|=4

=>x-1=4 hoặc x-1=-4

=>x=-3 hoặc x=5

b: =>|6x-5|=4

=>6x-5=4 hoặc 6x-5=-4

=>6x=1 hoặc 6x=9

=>x=1/6 hoặc x=3/2

Ta có  A > 0 

Từ đó  \(A^2=2+\sqrt{2+\sqrt{2+...}}\Leftrightarrow A^2=2+A\Leftrightarrow A^2-A-2=0\)

\(\Leftrightarrow\left(A+1\right)\left(A-2\right)=0\Leftrightarrow\orbr{\begin{cases}A+1=0\\A-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}A=-1\\A=2\end{cases}}\)

Do A > 0 nên A= 2

b, tương tự

c,\(C>2\)

Xét \(C^2=5+\sqrt{13+\sqrt{5+\sqrt{13...}}}\)

\(\left(C^2-5\right)^2=13+C\Leftrightarrow C^4-10C^2-C+12=0\Leftrightarrow\left(C^4-9C^2\right)-\left(C^2-9\right)-\left(C-3\right)=0\)

\(\Leftrightarrow\left(C-3\right)\left[\left(C+3\right)\left(C-1\right)\left(C+1\right)-1\right]=0\)

VÌ C> 2 =>  C-3 = 0 => C=3

à