Tính: làm chi tiết `\frac{2}{\sqrt{3}}“+“\frac{\sqrt{2}}{3}“+“\frac{2}{\sqrt{3}}“\sqrt{5/12-1/\sqrt{6}}`

Tính: làm chi tiết
`\frac{2}{\sqrt{3}}“+“\frac{\sqrt{2}}{3}“+“\frac{2}{\sqrt{3}}“\sqrt{5/12-1/\sqrt{6}}`

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  1. `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}`

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  2. 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}`

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