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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 9: Rút gọn: 1/ 2 + √3 + √2 / √6 – 2 / 3 + √3

Home/ Toán Lớp 9: Rút gọn: 1/ 2 + √3 + √2 / √6 – 2 / 3 + √3

Toán Lớp 9: Rút gọn: 1/ 2 + √3 + √2 / √6 – 2 / 3 + √3

Tháng Tám 12, 2022 Toán học 2 Comments 0 views

Toán Lớp 9: Rút gọn:
1/ 2 + √3 + √2 / √6 – 2 / 3 + √3

Comments ( 2 )

  1. dananhnhu2k4
    dananhnhu2k4
    12 Tháng Tám, 2022 at 2:52 sáng
    Reply
    Giải đáp: (3-sqrt3)/3 
    Lời giải và giải thích chi tiết:
    \(\dfrac{1}{2+\sqrt3}+\dfrac{\sqrt2}{\sqrt6} – \dfrac{2}{3+\sqrt3}\\=\dfrac{2-\sqrt3}{(2+\sqrt3)(2-\sqrt3)}+\dfrac{\sqrt2}{\sqrt2 . \sqrt3}-\dfrac{2(3-\sqrt3)}{(3+\sqrt3)(3-\sqrt3)}\\=\dfrac{2-\sqrt3}{1}+\dfrac{1}{\sqrt3}-\dfrac{2(3-\sqrt3)}{6}\\=2-\sqrt3+\dfrac{\sqrt3}{3}-\dfrac{3-\sqrt3}{3}=\dfrac{3(2-\sqrt3)+\sqrt3-(3-\sqrt3)}{3}=\dfrac{6-3\sqrt3+\sqrt3-3+\sqrt3}{3}=\dfrac{3-\sqrt3}{3}=\dfrac{\sqrt3.(\sqrt3-1)}{3}\\=\dfrac{\sqrt3-1}{\sqrt3}=\dfrac{\sqrt3(\sqrt3-1)}{3}=\dfrac{3-\sqrt3}{3}\)

  2. giahan44
    giahan44
    12 Tháng Tám, 2022 at 2:51 sáng
    Reply
    #tnvt
    \frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}
    =\frac{2-\sqrt{3}}{4-3}+\frac{\sqrt{2}}{\sqrt{3}.\sqrt{2}}-\frac{2(3-\sqrt{3})}{9-3}
    =2-\sqrt{3}+\frac{1}{\sqrt{3}}-\frac{3-\sqrt{3}}{3}
    =2-\sqrt{3}+\frac{1}{\sqrt{3}}-\frac{\sqrt{3}(\sqrt{3}-1)}{3}
    =2-\sqrt{3}+\frac{1-\sqrt{3}+1}{\sqrt{3}}
    =\frac{(2-\sqrt{3}).\sqrt{3}+2-\sqrt{3}}{\sqrt{3}}
    =\frac{2\sqrt{3}-3+2-\sqrt{3}}{\sqrt{3}}
    =\frac{\sqrt{3}-1}{\sqrt{3}}
    =\frac{3-\sqrt{3}}{3}
     

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222-9+11+12:2*14+14 = ? ( )

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