Tính: làm chi tiết `\frac{2}{\sqrt{3}}“+“\frac{\sqrt{2}}{3}“+“\frac{2}{\sqrt{3}}“\sqrt{5/12-1/\sqrt{6}}` 14/07/2021 Bởi Adalynn Tính: làm chi tiết `\frac{2}{\sqrt{3}}“+“\frac{\sqrt{2}}{3}“+“\frac{2}{\sqrt{3}}“\sqrt{5/12-1/\sqrt{6}}`
`2/\sqrt{3} +\sqrt{2}/3+ 2/\sqrt{3} . \sqrt{5/12-1/\sqrt{6}}` `=2\sqrt{3}/3+\sqrt{2}/3+2/\sqrt{3}. \sqrt{5/12-2\sqrt{6}/12}` `=(2\sqrt{3}+\sqrt{2})/3+2/\sqrt{3}. \sqrt{(5-2\sqrt{2.3})/12}` `=(2\sqrt{3}+\sqrt{2})/3+2/\sqrt{3}. \sqrt{(\sqrt{3}-\sqrt{2})^2/12}` `=(2\sqrt{3}+\sqrt{2})/3+2/\sqrt{3}. (\sqrt{3}-\sqrt{2})/(2\sqrt{3})` `=(2\sqrt{3}+\sqrt{2})/3+(2\sqrt{3}-2\sqrt{2})/6` `=(4\sqrt{3}+2\sqrt{2}+2\sqrt{3}-2\sqrt{2})/6` `=(6\sqrt{3})/6` `=\sqrt{3}` Bình luận
Đáp án: `2/sqrt{3}+sqrt{2}/3+2/sqrt{3}sqrt{5/12-1/\sqrt{6}}=sqrt{3}` Giải thích các bước giải: `2/sqrt{3}+sqrt{2}/3+2/sqrt{3}sqrt{5/12-1/\sqrt{6}}``=2/sqrt{3}+sqrt{2}/3+sqrt{(2/sqrt{3})^2 .(5/12-1/\sqrt{6})}``=2/sqrt{3}+sqrt{2}/3+sqrt{4/3 .(5/12-1/\sqrt{6})}``=2/sqrt{3}+sqrt{2}/3+sqrt{4/3 . 5/12-4/3 . 1/\sqrt{6})``=2/sqrt{3}+sqrt{2}/3+sqrt{5/9-(2sqrt{6})/9)``=2/sqrt{3}+sqrt{2}/3+sqrt{(5-2sqrt{6}). 1/9}``=2/sqrt{3}+sqrt{2}/3+sqrt{(5-2sqrt{6}). (1/3)^2}``=2/sqrt{3}+sqrt{2}/3+1/3sqrt{(5-2sqrt{6})}``=2/sqrt{3}+sqrt{2}/3+1/3sqrt{(sqrt{3})^2-2sqrt{2}.sqrt{3}+(sqrt{2})^2``=2/sqrt{3}+sqrt{2}/3+1/3sqrt{(sqrt{3}-sqrt{2})^2}``=(2sqrt{3})/3+sqrt{2}/3+1/3 .|sqrt{3}-sqrt{2}|``=(2sqrt{3})/3+sqrt{2}/3+1/3 .(sqrt{3}-sqrt{2})(text{vì 3>2 nên}sqrt{3}>sqrt{2})``=(2sqrt{3})/3+sqrt{2}/3+sqrt{3}/3 – sqrt{2}/3``=(2sqrt{3}+sqrt{2}+sqrt{3}-sqrt{2})/(3)``=(3sqrt{3})/3``=sqrt{3}` Bình luận
`2/\sqrt{3} +\sqrt{2}/3+ 2/\sqrt{3} . \sqrt{5/12-1/\sqrt{6}}`
`=2\sqrt{3}/3+\sqrt{2}/3+2/\sqrt{3}. \sqrt{5/12-2\sqrt{6}/12}`
`=(2\sqrt{3}+\sqrt{2})/3+2/\sqrt{3}. \sqrt{(5-2\sqrt{2.3})/12}`
`=(2\sqrt{3}+\sqrt{2})/3+2/\sqrt{3}. \sqrt{(\sqrt{3}-\sqrt{2})^2/12}`
`=(2\sqrt{3}+\sqrt{2})/3+2/\sqrt{3}. (\sqrt{3}-\sqrt{2})/(2\sqrt{3})`
`=(2\sqrt{3}+\sqrt{2})/3+(2\sqrt{3}-2\sqrt{2})/6`
`=(4\sqrt{3}+2\sqrt{2}+2\sqrt{3}-2\sqrt{2})/6`
`=(6\sqrt{3})/6`
`=\sqrt{3}`
Đáp án:
`2/sqrt{3}+sqrt{2}/3+2/sqrt{3}sqrt{5/12-1/\sqrt{6}}=sqrt{3}`
Giải thích các bước giải:
`2/sqrt{3}+sqrt{2}/3+2/sqrt{3}sqrt{5/12-1/\sqrt{6}}`
`=2/sqrt{3}+sqrt{2}/3+sqrt{(2/sqrt{3})^2 .(5/12-1/\sqrt{6})}`
`=2/sqrt{3}+sqrt{2}/3+sqrt{4/3 .(5/12-1/\sqrt{6})}`
`=2/sqrt{3}+sqrt{2}/3+sqrt{4/3 . 5/12-4/3 . 1/\sqrt{6})`
`=2/sqrt{3}+sqrt{2}/3+sqrt{5/9-(2sqrt{6})/9)`
`=2/sqrt{3}+sqrt{2}/3+sqrt{(5-2sqrt{6}). 1/9}`
`=2/sqrt{3}+sqrt{2}/3+sqrt{(5-2sqrt{6}). (1/3)^2}`
`=2/sqrt{3}+sqrt{2}/3+1/3sqrt{(5-2sqrt{6})}`
`=2/sqrt{3}+sqrt{2}/3+1/3sqrt{(sqrt{3})^2-2sqrt{2}.sqrt{3}+(sqrt{2})^2`
`=2/sqrt{3}+sqrt{2}/3+1/3sqrt{(sqrt{3}-sqrt{2})^2}`
`=(2sqrt{3})/3+sqrt{2}/3+1/3 .|sqrt{3}-sqrt{2}|`
`=(2sqrt{3})/3+sqrt{2}/3+1/3 .(sqrt{3}-sqrt{2})(text{vì 3>2 nên}sqrt{3}>sqrt{2})`
`=(2sqrt{3})/3+sqrt{2}/3+sqrt{3}/3 – sqrt{2}/3`
`=(2sqrt{3}+sqrt{2}+sqrt{3}-sqrt{2})/(3)`
`=(3sqrt{3})/3`
`=sqrt{3}`