Đáp án:
`D=\sqrt{3\sqrt{2}-4}`
Giải thích các bước giải:
Ta có:
`\qquad 27\sqrt{2}-38`
`=15\sqrt{2}+12\sqrt{2}-20-18`
`=5.3\sqrt{2}+4.3\sqrt{2}-5.4-(3\sqrt{2})^2`
`=5.3\sqrt{2}-(3\sqrt{2})^2-4.5+4.3\sqrt{2}`
`=3\sqrt{2}.(5-3\sqrt{2})-4.(5-3\sqrt{2})`
`=(5-3\sqrt{2}).(3\sqrt{2}-4)`
$\\$
`D= \sqrt{1+2\sqrt{27\sqrt{2}-38}}-\sqrt{5-3\sqrt{2}}`
`=\sqrt{5-3\sqrt{2}+2.\sqrt{(5-3\sqrt{2}).(3\sqrt{2}-4)}+3\sqrt{2}-4}-\sqrt{5-3\sqrt{2}}`
`=\sqrt{(\sqrt{5-3\sqrt{2}}+\sqrt{3\sqrt{2}-4})^2}-\sqrt{5-3\sqrt{2}}`
`=\sqrt{5-3\sqrt{2}}+\sqrt{3\sqrt{2}-4}-\sqrt{5-3\sqrt{2}}`
`=\sqrt{3\sqrt{2}-4}`
Vậy `D=\sqrt{3\sqrt{2}-4}`
