\(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

\(=\sqrt{\left(3\sqrt{5}\right)^2-2.3\sqrt{5}.1+1^2}-\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}.3+3^2}\)

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

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

\(\sqrt{\left(3\sqrt{5}\right)^2-2\cdot3\sqrt{5}\cdot1+1^2}-\sqrt{\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot3+3^2}\)

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

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

= \(3\sqrt{5}-1-2\sqrt{5}+3\)

= \(\sqrt{5}+2\)

Lời giải:

** Bạn chú ý lần sau ghi đầy đủ yêu cầu đề.

\(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}=\sqrt{45-2\sqrt{45}+1}-\sqrt{20-2\sqrt{20.9}+9}\)

\(=\sqrt{(\sqrt{45}-1)^2}-\sqrt{(\sqrt{20}-\sqrt{9})^2}=|\sqrt{45}-1|-|\sqrt{20}-\sqrt{9}|\)

\(=\sqrt{45}-1-(\sqrt{20}-\sqrt{9})=3\sqrt{5}-1-2\sqrt{5}+3=\sqrt{5}+2\)

a) \(\sqrt{6-4\sqrt{2}}+\sqrt{19-6\sqrt{2}}\)\(=\sqrt{4-4\sqrt{2}+2}+\sqrt{18-2.3\sqrt{2}.1+1}=\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-1\right)^2}\)\(=\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-1\right)^2}\)

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

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

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

b) \(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

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

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

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

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

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

c) \(\sqrt{7-2\sqrt{10}}-\sqrt{6-2\sqrt{5}}\)

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

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

= / \(\sqrt{5}\) - \(\sqrt{2}\) / - / \(\sqrt{5}\) - 1 /

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

a) \(\sqrt{6-4\sqrt{2}}+\sqrt{19-6\sqrt{2}}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

a, \(\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)

= \(\left|\text{√}5-\text{√}2\right|\)-\(\left|\text{√}5+\text{√}2\right|\)

= √5 -√2 - √5 - √2

= -2√2

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

= \(\left|1-\text{√}2\right|\) - \(\left|2-\text{√}5\right|\)

= √2 - 1 + 2 - √5

= √2-√5 +1

c, \(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

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

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

= 3√5 - 1 - 2√5 +3

= √5 + 2

Mình làm luôn nhé :

\(\sqrt{45-2.3\sqrt{5}+1}-\sqrt{20-2.3.2\sqrt{5}+9}\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5-\sqrt{45+2.2.\sqrt{2}.3\sqrt{5}+8}}\left(\sqrt{3}+\sqrt{5}\right).\sqrt{5-2.\sqrt{5}.\sqrt{2}+2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{7+2.\sqrt{7}.\sqrt{3}+3}\) Tới đây dễ rồi , bạn tự nhóm HĐT là ra ::v

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

   2, (tương tự ý 1 cũng tách thành hằng đẳng thức \(\sqrt{46-6\sqrt{5}}=\sqrt{\left(3\sqrt{5}-1\right)^2}\)và \(\sqrt{29-12\sqrt{5}}=\sqrt{\left(2\sqrt{5}-3\right)^2}\)

    3,\(\left(\sqrt{2}-\sqrt{9}\right)\left(\sqrt{\left(3+\sqrt{2}\right)^2}\right)=\left(\sqrt{2}-3\right)\left(\sqrt{2}+3\right)=2-9=-7\)

4, tương tự ý 3

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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