Rút gọn a) 9- √17 – √9+ √17 b) √11-2√30 + √13+2√42 mình cần gấp cảm ơn mn hứa 5* và TLHN. 30/06/2021 Bởi aikhanh Rút gọn a) 9- √17 – √9+ √17 b) √11-2√30 + √13+2√42 mình cần gấp cảm ơn mn hứa 5* và TLHN.
Đáp án: `a)sqrt{9-sqrt{17}}-sqrt{9+sqrt{17}}` `=sqrt{(18-2sqrt{17})/2}-sqrt{(18+2sqrt{17})/2}` `=sqrt{(17-2sqrt{17}+1)/2}-sqrt{(17+2sqrt{17}+1)/2}` `=sqrt{(sqrt{17}-1)^2/2}-sqrt{(sqrt{17}+1)^2/2}` `=(sqrt{17}-1)/sqrt2-(sqrt{17}+1)/sqrt2` `=(sqrt{17}-1-sqrt{17}-1)/sqrt2` `=-sqrt2` `b)sqrt{11-2sqrt{30}}+sqrt{13+2sqrt{42}}` `=sqrt{6-2sqrt{6}.sqrt5+5}+sqrt{6+2sqrt{6}.sqrt7+7}` `=sqrt{(sqrt6-sqrt5)^2}+sqrt{(sqrt6+sqrt7)^2}` `=|sqrt6-sqrt5|+|sqrt6+sqrt7|` `=sqrt6-sqrt5+sqrt6+sqrt7` `=2sqrt6+sqrt7-sqrt5` Bình luận
Hướng dẫn trả lời: $\text{a) $\sqrt{9 – \sqrt{17}}$ – $\sqrt{9 + \sqrt{17}}$.}$ $\text{Đặt N = $\sqrt{9 – \sqrt{17}}$ – $\sqrt{9 + \sqrt{17}}$.}$ $\text{⇒ $N^2$ = $(\sqrt{9 – \sqrt{17}}$ – $\sqrt{9 + \sqrt{17}})^2$.}$ $\text{$N^2$ = $9 – \sqrt{17} – 2.\sqrt{9 – \sqrt{17}}.\sqrt{9 + \sqrt{17}} + 9 + \sqrt{17}$.}$ $\text{$N^2$ = $18 – 2.\sqrt{(9 – \sqrt{17}).(9 + \sqrt{17})}$.}$ $\text{$N^2$ = $18 – 2.\sqrt{81 – 17}$.}$ $\text{$N^2$ = $18 – 2.\sqrt{64}$.}$ $\text{$N^2$ = $18 – 2.8$.}$ $\text{$N^2$ = $18 – 16$.}$ $\text{$N^2$ = 2.}$ $\text{→ N = ± $\sqrt{2}$.}$ $\text{Mà N < 0 (Vì $\sqrt{9 – \sqrt{17}}$ < $\sqrt{9 + \sqrt{17}}$).}$ Nên N = – $\sqrt{2}$. $\text{Vậy giá trị của $\sqrt{9 – \sqrt{17}}$ – $\sqrt{9 + \sqrt{17}}$ là}$ – $\sqrt{2}$. $\text{b) $\sqrt{11 – 2\sqrt{30}}$ + $\sqrt{13 + 2\sqrt{42}}$.}$ Ta có: $\sqrt{11 – 2\sqrt{30}}$ = $\sqrt{6 – 2\sqrt{6}.\sqrt{5} + 5}$ = $\sqrt{(\sqrt{6}- \sqrt{5})^2}$ = $|\sqrt{6} – \sqrt{5}|$ = $\sqrt{6} – \sqrt{5}$. (Vì $\sqrt{6} – \sqrt{5}$ > 0) Và: $\sqrt{13 + 2\sqrt{42}}$ = $\sqrt{7 + 2\sqrt{7}.\sqrt{6} + 6}$ = $\sqrt{(\sqrt{7}+ \sqrt{6})^2}$ = $|\sqrt{7} + \sqrt{6}|$ = $\sqrt{7} + \sqrt{6}$. (Vì $\sqrt{7} + \sqrt{6}$ > 0) $\text{Nên $\sqrt{11 – 2\sqrt{30}}$ + $\sqrt{13 + 2\sqrt{42}}$.}$ $\text{= ($\sqrt{6} – \sqrt{5}$) + ($\sqrt{7} + \sqrt{6}$).}$ $\text{= $\sqrt{6} – \sqrt{5}$ + $\sqrt{7} + \sqrt{6}$.}$ $\text{= $2\sqrt{6} – \sqrt{5} + \sqrt{7}$.}$ Bình luận
Đáp án:
`a)sqrt{9-sqrt{17}}-sqrt{9+sqrt{17}}`
`=sqrt{(18-2sqrt{17})/2}-sqrt{(18+2sqrt{17})/2}`
`=sqrt{(17-2sqrt{17}+1)/2}-sqrt{(17+2sqrt{17}+1)/2}`
`=sqrt{(sqrt{17}-1)^2/2}-sqrt{(sqrt{17}+1)^2/2}`
`=(sqrt{17}-1)/sqrt2-(sqrt{17}+1)/sqrt2`
`=(sqrt{17}-1-sqrt{17}-1)/sqrt2`
`=-sqrt2`
`b)sqrt{11-2sqrt{30}}+sqrt{13+2sqrt{42}}`
`=sqrt{6-2sqrt{6}.sqrt5+5}+sqrt{6+2sqrt{6}.sqrt7+7}`
`=sqrt{(sqrt6-sqrt5)^2}+sqrt{(sqrt6+sqrt7)^2}`
`=|sqrt6-sqrt5|+|sqrt6+sqrt7|`
`=sqrt6-sqrt5+sqrt6+sqrt7`
`=2sqrt6+sqrt7-sqrt5`
Hướng dẫn trả lời:
$\text{a) $\sqrt{9 – \sqrt{17}}$ – $\sqrt{9 + \sqrt{17}}$.}$
$\text{Đặt N = $\sqrt{9 – \sqrt{17}}$ – $\sqrt{9 + \sqrt{17}}$.}$
$\text{⇒ $N^2$ = $(\sqrt{9 – \sqrt{17}}$ – $\sqrt{9 + \sqrt{17}})^2$.}$
$\text{$N^2$ = $9 – \sqrt{17} – 2.\sqrt{9 – \sqrt{17}}.\sqrt{9 + \sqrt{17}} + 9 + \sqrt{17}$.}$
$\text{$N^2$ = $18 – 2.\sqrt{(9 – \sqrt{17}).(9 + \sqrt{17})}$.}$
$\text{$N^2$ = $18 – 2.\sqrt{81 – 17}$.}$
$\text{$N^2$ = $18 – 2.\sqrt{64}$.}$
$\text{$N^2$ = $18 – 2.8$.}$
$\text{$N^2$ = $18 – 16$.}$
$\text{$N^2$ = 2.}$
$\text{→ N = ± $\sqrt{2}$.}$
$\text{Mà N < 0 (Vì $\sqrt{9 – \sqrt{17}}$ < $\sqrt{9 + \sqrt{17}}$).}$
Nên N = – $\sqrt{2}$.
$\text{Vậy giá trị của $\sqrt{9 – \sqrt{17}}$ – $\sqrt{9 + \sqrt{17}}$ là}$ – $\sqrt{2}$.
$\text{b) $\sqrt{11 – 2\sqrt{30}}$ + $\sqrt{13 + 2\sqrt{42}}$.}$
Ta có: $\sqrt{11 – 2\sqrt{30}}$ = $\sqrt{6 – 2\sqrt{6}.\sqrt{5} + 5}$ = $\sqrt{(\sqrt{6}- \sqrt{5})^2}$ = $|\sqrt{6} – \sqrt{5}|$ = $\sqrt{6} – \sqrt{5}$. (Vì $\sqrt{6} – \sqrt{5}$ > 0)
Và: $\sqrt{13 + 2\sqrt{42}}$ = $\sqrt{7 + 2\sqrt{7}.\sqrt{6} + 6}$ = $\sqrt{(\sqrt{7}+ \sqrt{6})^2}$ = $|\sqrt{7} + \sqrt{6}|$ = $\sqrt{7} + \sqrt{6}$. (Vì $\sqrt{7} + \sqrt{6}$ > 0)
$\text{Nên $\sqrt{11 – 2\sqrt{30}}$ + $\sqrt{13 + 2\sqrt{42}}$.}$
$\text{= ($\sqrt{6} – \sqrt{5}$) + ($\sqrt{7} + \sqrt{6}$).}$
$\text{= $\sqrt{6} – \sqrt{5}$ + $\sqrt{7} + \sqrt{6}$.}$
$\text{= $2\sqrt{6} – \sqrt{5} + \sqrt{7}$.}$