Đáp án:
`a)` `{\sqrt{2}.(\sqrt{7}+1)}/2`
`b)` `\sqrt{6}/2`
Giải thích các bước giải:
`a)` `\sqrt{\sqrt{7}+5-2\sqrt{\sqrt{7}+4}}+1`
`=\sqrt{\sqrt{7}+4-2.\sqrt{\sqrt{7}+4}+1}+1`
`=\sqrt{(\sqrt{\sqrt{7}+4}-1)^2}+1`
`=|\sqrt{\sqrt{7}+4}-1|+1`
`=\sqrt{\sqrt{7}+4}-1+1`
`=\sqrt{\sqrt{7}+4}=\sqrt{1/2 .(8+2\sqrt{7})}`
`=\sqrt{1/2 .(7+2.\sqrt{7}.1+1)}`
`=\sqrt{1/2 .(\sqrt{7}+1)^2}`
`=|1/\sqrt{2} . (\sqrt{7}+1)|`
`={\sqrt{7}+1}/\sqrt{2}`
`={\sqrt{2} . (\sqrt{7}+1)}/2`
Vậy: `\sqrt{\sqrt{7}+5-2\sqrt{\sqrt{7}+4}}+1={\sqrt{2}.(\sqrt{7}+1)}/2`
$\\$
`b)` `\sqrt{4+\sqrt{3}+\sqrt{6\sqrt{3}+15}}-\sqrt{\sqrt{3}+5/2}`
`=\sqrt{1/ 2 . [8+2\sqrt{3}+2\sqrt{3.(2\sqrt{3}+5)}]}-\sqrt{\sqrt{3}+5/2}`
`=\sqrt{1/ 2 . (3+2.\sqrt{3}.\sqrt{2\sqrt{3}+5}+2\sqrt{3}+5)}-\sqrt{\sqrt{3}+5/2}`
`=\sqrt{1/ 2 .(\sqrt{3}+\sqrt{2\sqrt{3}+5})^2}-\sqrt{\sqrt{3}+5/2}`
`=|1/\sqrt{2} . (\sqrt{3}+\sqrt{2\sqrt{3}+5})|-\sqrt{{2\sqrt{3}+5}/{2}}`
`=1/{\sqrt{2}}. (\sqrt{3}+\sqrt{2\sqrt{3}+5})- \sqrt{2\sqrt{3}+5}/{\sqrt{2}}`
`=\sqrt{3}/{\sqrt{2}}+\sqrt{2\sqrt{3}+5}/{\sqrt{2}}-\sqrt{2\sqrt{3}+5}/{\sqrt{2}}`
`=\sqrt{3}/\sqrt{2}={\sqrt{6}}/2`
Vậy: `\sqrt{4+\sqrt{3}+\sqrt{6\sqrt{3}+15}}-\sqrt{\sqrt{3}+5/2}=\sqrt{6}/2`
