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\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)= -8,94427191 NHOA! Nguyễn Diễm Quỳnh

K VÀ KB NHOA ! 

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

\(\Leftrightarrow B=\sqrt{13-4\sqrt{10}}-\sqrt{53+12\sqrt{10}}\)

\(\Leftrightarrow B^2=66+8\sqrt{10}-2.\sqrt{13-4\sqrt{10}}.\sqrt{53+12\sqrt{10}}\)

\(=66+8\sqrt{10}-2.\sqrt{209-56\sqrt{10}}\)

\(=66+8\sqrt{10}-2.\sqrt{\left(4\sqrt{10}-7\right)^2}\)

\(=66+8\sqrt{10}-8\sqrt{10}+14=80\)

\(\Rightarrow B=-\sqrt{80}=-4\sqrt{5}\)

\(D=\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}=\sqrt{13-4\sqrt{10}}-\sqrt{53+12\sqrt{10}}\)

\(=\sqrt{8-2.2\sqrt{2}\sqrt{5}+5}-\sqrt{\text{coi lại đề}}\)

\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{60}}=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(4\sqrt{3}+\sqrt{5}\right)^2}\)

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

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

\(\sqrt{\left(4+\sqrt{3}\right)\sqrt{19-8\sqrt{3}}+3}=\sqrt{\left(4+\sqrt{3}\right)\sqrt{\left(4-\sqrt{3}\right)^2}+3}\)

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

a) Ta có: \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{60}}\)

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

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

b) Ta có: \(\sqrt{\left(4+\sqrt{3}\right)\cdot\sqrt{19-8\sqrt{3}+3}}\)

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

=4

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

\(=\sqrt{13-4\sqrt{10}}+\sqrt{53+12\sqrt{10}}\)

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

\(=2\sqrt{5}\)

`\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}`

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

`=|2\sqrt{2}-\sqrt{5}|+3\sqrt{5}+2\sqrt{2}`

`=2\sqrt{2}-\sqrt{5}+3\sqrt{5}+2\sqrt{2}`

`=4\sqrt{2}+2\sqrt{5}`

\(=\sqrt{13-4\sqrt{10}}-\sqrt{53-12\sqrt{10}}\)

\(=\sqrt{13-2\cdot2\sqrt{2}\cdot\sqrt{5}}-\sqrt{53-2\cdot3\sqrt{5}\cdot2\sqrt{2}}\)

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

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

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

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

\(=\sqrt{13-\sqrt{4^2\cdot10}}-\sqrt{53+4\sqrt{3^2\cdot10}}\)

\(=\sqrt{13-4\sqrt{10}}-\sqrt{53+12\sqrt{10}}\)

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

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

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

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

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

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

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

\(=\sqrt{13-2\sqrt{40}}-\sqrt{53+2\sqrt{360}}\)

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

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

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

\(=\sqrt{8}-\sqrt{5}-3\sqrt{5}-\sqrt{8}\)

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

\(\left(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\right)\left(\sqrt{10}-\sqrt{2}\right)\)

\(=2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\times\sqrt{2}\left(\sqrt{5}-1\right)\)

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

\(=2\sqrt{6+2\sqrt{5}}\times\left(\sqrt{5}-\sqrt{1}\right)\)

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

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

\(=2\left(5-1\right)\)

= 8

~ ~ ~

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

\(=\sqrt{13-4\sqrt{10}}-\sqrt{53+12\sqrt{10}}\)

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

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

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

a. \(\left(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\right)\left(\sqrt{10}-\sqrt{2}\right)=\left[2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\right]\left(\sqrt{10}-\sqrt{2}\right)=\left(2\sqrt{4+\sqrt{5}-1}\right)\left(\sqrt{10}-\sqrt{2}\right)=\left(2\sqrt{3+\sqrt{5}}\right)\left(\sqrt{10}-\sqrt{2}\right)=\left[2\sqrt{\left(\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{1}{2}}\right)^2}\right]\left(\sqrt{10}-\sqrt{2}\right)=\left[2\left(\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{1}{2}}\right)\right]\left(\sqrt{10}-\sqrt{2}\right)=\left(\sqrt{10}+\sqrt{2}\right)\left(\sqrt{10}-\sqrt{2}\right)=10-2=8\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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