Tính: làm chi tiết giúp em `\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}“+“\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}` 14/07/2021 Bởi Elliana Tính: làm chi tiết giúp em `\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}“+“\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}`
$\dfrac{2+\sqrt3}{\sqrt2+\sqrt{2+\sqrt3}}+\dfrac{2-\sqrt3}{\sqrt2-\sqrt{2-\sqrt3}}$ Nhân cả tử và mẫu hai hạng tử với $\sqrt2$: $\dfrac{2\sqrt2+\sqrt6}{2+\sqrt{4+2\sqrt3}}+\dfrac{2\sqrt2-\sqrt6}{2-\sqrt{4-2\sqrt3}}$ $=\dfrac{2\sqrt2+\sqrt6}{2+\sqrt{(\sqrt3+1)^2}}+\dfrac{2\sqrt2-\sqrt6}{2-\sqrt{(\sqrt3-1)^2}}$ $=\dfrac{\sqrt2(2+\sqrt3)}{2+\sqrt3+1}+\dfrac{\sqrt2(2-\sqrt3)}{2-\sqrt3+1}$ $=\dfrac{\sqrt2(2+\sqrt3)}{3+\sqrt3}+\dfrac{\sqrt2(2-\sqrt3)}{3-\sqrt3}$ $=\dfrac{\sqrt2(2+\sqrt3)(3-\sqrt3)+\sqrt2(2-\sqrt3)(3+\sqrt3)}{9-3}$ $=\dfrac{\sqrt2(6-2\sqrt3+3\sqrt3-3+6+2\sqrt3-3\sqrt3-3)}{6}$ $=\dfrac{\sqrt2(12-6)}{6}$ $=\sqrt2$ Bình luận
Của em đây nhé! `(2+sqrt3)/(sqrt2+sqrt{2+sqrt3})+(2-sqrt3)/(sqrt2-sqrt{2-sqrt3})` `=(2sqrt2+sqrt6)/(sqrt2.sqrt2+sqrt2.sqrt{2+sqrt3})+(2sqrt2-sqrt6)/(sqrt2.sqrt2-sqrt2.sqrt{2+sqrt3})` `=(2sqrt2+sqrt6)/(2+sqrt{4+2sqrt3})+(2sqrt2-sqrt6)/(2-sqrt{4-2sqrt3})` `=(2sqrt2+sqrt6)/(2+sqrt{3+2sqrt3+1})+(2sqrt2-sqrt6)/(2-sqrt{3-2sqrt3+1})` `=(2sqrt2+sqrt6)/(2+sqrt{(sqrt3+1)^2})+(2sqrt2-6)/(2-sqrt{(sqrt3-1)^2})` `=(2sqrt2+sqrt6)/(2+sqrt3+1)+(2sqrt2-sqrt6)/(2-sqrt3+1)` `=(2sqrt2+sqrt6)/(3+sqrt3)+(2sqrt2-sqrt6)/(3-sqrt3)` `=((2sqrt2+sqrt6)(3-sqrt3))/(9-3)+((2sqrt2-sqrt6)(3+sqrt3))/(9-3)` `=((2sqrt2+sqrt6)(3-sqrt3)+(2sqrt2-sqrt6)(3+sqrt3))/6` `=(6sqrt2-2sqrt6+3sqrt6-sqrt{18}+6sqrt2+2sqrt6-3sqrt6-sqrt{18})/6` `=(12sqrt2-2sqrt{18})/6` `=(12sqrt2-2.sqrt{9.2})/6` `=(12sqrt2-6sqrt2)/6` `=(6sqrt2)/6` `=sqrt2`. Bình luận
$\dfrac{2+\sqrt3}{\sqrt2+\sqrt{2+\sqrt3}}+\dfrac{2-\sqrt3}{\sqrt2-\sqrt{2-\sqrt3}}$
Nhân cả tử và mẫu hai hạng tử với $\sqrt2$:
$\dfrac{2\sqrt2+\sqrt6}{2+\sqrt{4+2\sqrt3}}+\dfrac{2\sqrt2-\sqrt6}{2-\sqrt{4-2\sqrt3}}$
$=\dfrac{2\sqrt2+\sqrt6}{2+\sqrt{(\sqrt3+1)^2}}+\dfrac{2\sqrt2-\sqrt6}{2-\sqrt{(\sqrt3-1)^2}}$
$=\dfrac{\sqrt2(2+\sqrt3)}{2+\sqrt3+1}+\dfrac{\sqrt2(2-\sqrt3)}{2-\sqrt3+1}$
$=\dfrac{\sqrt2(2+\sqrt3)}{3+\sqrt3}+\dfrac{\sqrt2(2-\sqrt3)}{3-\sqrt3}$
$=\dfrac{\sqrt2(2+\sqrt3)(3-\sqrt3)+\sqrt2(2-\sqrt3)(3+\sqrt3)}{9-3}$
$=\dfrac{\sqrt2(6-2\sqrt3+3\sqrt3-3+6+2\sqrt3-3\sqrt3-3)}{6}$
$=\dfrac{\sqrt2(12-6)}{6}$
$=\sqrt2$
Của em đây nhé!
`(2+sqrt3)/(sqrt2+sqrt{2+sqrt3})+(2-sqrt3)/(sqrt2-sqrt{2-sqrt3})`
`=(2sqrt2+sqrt6)/(sqrt2.sqrt2+sqrt2.sqrt{2+sqrt3})+(2sqrt2-sqrt6)/(sqrt2.sqrt2-sqrt2.sqrt{2+sqrt3})`
`=(2sqrt2+sqrt6)/(2+sqrt{4+2sqrt3})+(2sqrt2-sqrt6)/(2-sqrt{4-2sqrt3})`
`=(2sqrt2+sqrt6)/(2+sqrt{3+2sqrt3+1})+(2sqrt2-sqrt6)/(2-sqrt{3-2sqrt3+1})`
`=(2sqrt2+sqrt6)/(2+sqrt{(sqrt3+1)^2})+(2sqrt2-6)/(2-sqrt{(sqrt3-1)^2})`
`=(2sqrt2+sqrt6)/(2+sqrt3+1)+(2sqrt2-sqrt6)/(2-sqrt3+1)`
`=(2sqrt2+sqrt6)/(3+sqrt3)+(2sqrt2-sqrt6)/(3-sqrt3)`
`=((2sqrt2+sqrt6)(3-sqrt3))/(9-3)+((2sqrt2-sqrt6)(3+sqrt3))/(9-3)`
`=((2sqrt2+sqrt6)(3-sqrt3)+(2sqrt2-sqrt6)(3+sqrt3))/6`
`=(6sqrt2-2sqrt6+3sqrt6-sqrt{18}+6sqrt2+2sqrt6-3sqrt6-sqrt{18})/6`
`=(12sqrt2-2sqrt{18})/6`
`=(12sqrt2-2.sqrt{9.2})/6`
`=(12sqrt2-6sqrt2)/6`
`=(6sqrt2)/6`
`=sqrt2`.