sai de thi phai

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

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

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

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

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

\(=2\sqrt{5}\)

`\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}`

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Bài này không sai đề , tớ làm lại cho :

\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}=\sqrt{13-4\sqrt{10}}-\sqrt{53+12\sqrt{10}}=\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}-\sqrt{45+2.3\sqrt{5}.2\sqrt{2}+8}=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}=\text{ |}2\sqrt{2}-\sqrt{5}\text{ |}-\text{ |}3\sqrt{5}+2\sqrt{2}\text{ |}=4\sqrt{2}-4\sqrt{5}\)

Đề này mình làm không ra nên mình sẽ sửa đề.

Giải:

\(\sqrt{14-\sqrt{160}}-\sqrt{49+4\sqrt{90}}\)

\(=\sqrt{14-4\sqrt{10}}-\sqrt{49+12\sqrt{10}}\)

\(=\sqrt{10-4\sqrt{10}+4}-\sqrt{40+12\sqrt{10}+9}\)

\(=\sqrt{\left(\sqrt{10}\right)^2-2.\sqrt{10}.2+2^2}-\sqrt{\left(2\sqrt{10}\right)^2+2.2\sqrt{10}.3+3^2}\)

\(=\sqrt{\left(\sqrt{10}-2\right)^2}-\sqrt{\left(2\sqrt{10}+3\right)^2}\)

\(=\sqrt{10}-2-\left(2\sqrt{10}+3\right)\)

\(=\sqrt{10}-2-2\sqrt{10}-3\)

\(=-\sqrt{10}-5\)

Vậy ...

Nếu sai mong bạn thông cảm

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

\(=\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}-\sqrt{45+2.3\sqrt{5}.2\sqrt{2}+8}\)

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

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

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

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

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

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

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

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

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

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