Đáp án:
`b)` `18`
`c)` `{\sqrt{6}+\sqrt{2}}/2`
`d)` ` 5+3\sqrt{2}`
Giải thích các bước giải:
`b)` `(4-\sqrt{7})(\sqrt{2}+\sqrt{14}).\sqrt{4+\sqrt{7}} `
`=(\sqrt{2}+\sqrt{14}).\sqrt{4-\sqrt{7}}.\sqrt{4-\sqrt{7}}.\sqrt{4+\sqrt{7}}`
`=(1+\sqrt{7}).\sqrt{2}.\sqrt{4-\sqrt{7}}.\sqrt{(4-\sqrt{7}).(4+\sqrt{7})}`
`=(1+\sqrt{7}).\sqrt{8-2\sqrt{7}}.\sqrt{4^2-7}`
`=(1+\sqrt{7}).\sqrt{7-2\sqrt{7}.1+1^2}.\sqrt{9}`
`=(1+\sqrt{7}).\sqrt{(\sqrt{7}-1)^2}.3`
`=3(1+\sqrt{7}).|\sqrt{7}-1|`
`=3(1+\sqrt{7}).(\sqrt{7}-1)`
`=3.(7-1^2)=3.6=18`
Vậy: `(4-\sqrt{7})(\sqrt{2}+\sqrt{14}).\sqrt{4+\sqrt{7}}=18 `
$\\$
`c)` `\sqrt{4+\sqrt{15}}-\sqrt{4-\sqrt{15}}-\sqrt{2-\sqrt{3}}`
`=\sqrt{{8+2\sqrt{15}}/2}-\sqrt{{8-2\sqrt{15}}/2}-\sqrt{{4-2\sqrt{3}}/2}`
`=\sqrt{{5+2\sqrt{5}.\sqrt{3}+3}/2}-\sqrt{{5-2\sqrt{5}.\sqrt{3}+3}/2}-\sqrt{{3-2\sqrt{3}.1+1^2}/2}`
`=\sqrt{{(\sqrt{5}+\sqrt{3})^2}/2}-\sqrt{{(\sqrt{5}-\sqrt{3})^2}/2}-\sqrt{{(\sqrt{3}-1)^2}/2}`
`=|{\sqrt{5}+\sqrt{3}}/\sqrt{2}|-|{\sqrt{5}-\sqrt{3}}/\sqrt{2}|-|{\sqrt{3}-1}/\sqrt{2}|`
`={\sqrt{5}+\sqrt{3}}/\sqrt{2}-{\sqrt{5}-\sqrt{3}}/\sqrt{2}-{\sqrt{3}-1}/\sqrt{2}`
`={\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}-\sqrt{3}+1}/\sqrt{2}`
`={\sqrt{3}+1}/\sqrt{2}={\sqrt{6}+\sqrt{2}}/2`
Vậy: `\sqrt{4+\sqrt{15}}-\sqrt{4-\sqrt{15}}-\sqrt{2-\sqrt{3}}={\sqrt{6}+\sqrt{2}}/2`
$\\$
`d)` `\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}`
`=\sqrt{13+30\sqrt{2+\sqrt{(2\sqrt{2})^2+2.2\sqrt{2}.1+1^2}}}`
`=\sqrt{13+30\sqrt{2+\sqrt{(2\sqrt{2}+1)^2}}}`
`=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}`
`=\sqrt{13+30\sqrt{(\sqrt{2}+1)^2}}`
`=\sqrt{13+30(\sqrt{2}+1)}`
`=\sqrt{43+30\sqrt{2}}=\sqrt{25+30\sqrt{2}+18}`
`=\sqrt{5^2+2.5.3\sqrt{2}+(3\sqrt{2})^2}`
`=\sqrt{(5+3\sqrt{2})^2}=|5+3\sqrt{2}|=5+3\sqrt{2}`
Vậy: `\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=5+3\sqrt{2}`
