\(a.\sqrt{8-\sqrt{28}}+\sqrt{21+12\sqrt{3}}=\sqrt{7-2\sqrt{7}+1}+\sqrt{12+2.2\sqrt{3}.3+9}=\sqrt{7}-1+2\sqrt{3}+3=2\sqrt{3}+\sqrt{7}+2\) \(b.\sqrt{5+\sqrt{24}}-\sqrt{57-40\sqrt{2}}=\sqrt{3+2.\sqrt{3}.\sqrt{2}+2}-\sqrt{32-2.4\sqrt{2}.5+25}=\sqrt{3}+\sqrt{2}-4\sqrt{2}+5=\sqrt{3}-3\sqrt{2}+5\) \(c.\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}=\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}+\sqrt{45+2.3\sqrt{5}.2\sqrt{2}+8}=2\sqrt{2}-\sqrt{5}+3\sqrt{5}+2\sqrt{2}=4\sqrt{2}+2\sqrt{5}\)

Bài 1:

a) Ta có: \(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

\(=\sqrt{45-2\cdot\sqrt{45}\cdot1+1}-\sqrt{9-2\cdot\sqrt{9}\cdot\sqrt{20}+20}\)

\(=\sqrt{\left(\sqrt{45}-1\right)^2}-\sqrt{\left(3-\sqrt{20}\right)^2}\)

\(=\left|\sqrt{45}-1\right|-\left|3-\sqrt{20}\right|\)

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

\(=\sqrt{45}+\sqrt{20}-4\)

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

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

\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{8}+8}-\sqrt{45+2\cdot\sqrt{45}\cdot\sqrt{8}+8}\)

\(=\sqrt{\left(\sqrt{5}-\sqrt{8}\right)^2}-\sqrt{\left(\sqrt{45}+\sqrt{8}\right)^2}\)

\(=\left|\sqrt{5}-\sqrt{8}\right|-\left|\sqrt{45}+\sqrt{8}\right|\)

\(=\sqrt{8}-\sqrt{5}-\sqrt{45}-\sqrt{8}\)

\(=-\sqrt{5}-\sqrt{45}=-\sqrt{5}\left(1+\sqrt{9}\right)=-4\sqrt{5}\)

c) Ta có: \(\left(3-\sqrt{2}\right)\cdot\sqrt{7+4\sqrt{3}}\)

\(=\left(3-\sqrt{2}\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot2+4}\)

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

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

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

d) Ta có: \(\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10+2\sqrt{21}}\)

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

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

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

\(=\left(\sqrt{7}\right)^2-\left(\sqrt{3}\right)^2=7-3=4\)

à

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

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

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

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

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

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

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

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

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

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

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

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

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

* \(\sqrt{2}\)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)~~~~~

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

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

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

\(=2.\sqrt{\frac{1}{2}}=\sqrt{2}\)

*****b)*****

(Hình như đề có cái gì đó sai sai hả bạn?)

~~~~~c)~~~~~

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

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

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

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

\(=-2\)

*****d)*****

     \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)

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

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

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

(Chúc bạn học tốt và tíck cho mìk vs nhé ~~~~~bạn xem lại câu b hộ mình luôn nha~~~~~!)

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