Những câu hỏi liên quan
Minh_28_Anh_09_Lê
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Hoàng Linh Chi
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Hoàng Tử Hà
17 tháng 6 2019 lúc 19:54

a/ \(A=\frac{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}{2-\sqrt{3}}+\frac{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}{2+\sqrt{3}}\)

\(A=\frac{2+\sqrt{3}+2-\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{4}{1}=4\)

b/\(A=\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\frac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}\)

\(A=\frac{\sqrt{2}-1}{3-2\sqrt{2}}-\frac{\sqrt{2}+1}{3+2\sqrt{2}}\)

\(A=\frac{\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)-\left(\sqrt{2}+1\right)\left(3-2\sqrt{2}\right)}{9-8}\)

\(A=3\sqrt{2}+4-3-2\sqrt{2}-3\sqrt{2}+4-3+2\sqrt{2}=8\)

c/ \(A=\frac{\left(\sqrt{5}+\sqrt{3}\right)^2+\left(\sqrt{5}-\sqrt{3}\right)^2}{5-3}\)

\(A=\frac{5+2\sqrt{15}+3+5-2\sqrt{15}+3}{2}=8\)

d/ theo câu c có \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=8\)

\(\Rightarrow A=8-\frac{\left(\sqrt{5}+1\right)^2}{5-1}=\frac{32-5-2\sqrt{5}-1}{4}=\frac{2\left(13-\sqrt{5}\right)}{4}=\frac{13-\sqrt{5}}{2}\)

Hỏi Làm Gì
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RI AM
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Nobi Nobita
18 tháng 10 2020 lúc 20:23

\(\frac{\sqrt{\sqrt{5}+\sqrt{2}}}{\sqrt{3\sqrt{5}-3\sqrt{2}}}=\frac{\sqrt{\sqrt{5}+\sqrt{2}}}{\sqrt{3.\left(\sqrt{5}-\sqrt{2}\right)}}=\frac{\sqrt{\sqrt{5}+\sqrt{2}}}{\sqrt{3}.\sqrt{\sqrt{5}-\sqrt{2}}}\)

\(=\frac{(\sqrt{\sqrt{5}+\sqrt{2}})^2}{\sqrt{3}.\sqrt{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}}=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{3}.\sqrt{5-2}}\)

\(=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{3}.\sqrt{3}}=\frac{\sqrt{5}+\sqrt{2}}{3}\)

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Lưu Như Ý
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Park Chanyeol
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Ngọc Vĩ
16 tháng 7 2016 lúc 19:38

a/ Bạn ghi nhầm đề rồi

c/ \(2\sqrt{18\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5\sqrt{48}}\)   

     \(=2\sqrt{18}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{\sqrt{48}}\)

       \(=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{4\sqrt{3}}\)

       \(=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-6\sqrt{5}.\sqrt{\sqrt{3}}\)

        \(=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-\sqrt{5}-3\sqrt{5}\right)\)

         \(=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-4\sqrt{5}\right)\)\(=2\sqrt{2\sqrt{3}}\left(3-2\sqrt{10}\right)\)

f/ \(\sqrt{2}.\sqrt{2+\sqrt{3}}-2\left(\sqrt{3}-1\right)=\sqrt{4+2\sqrt{3}}-2\left(\sqrt{3}-1\right)\)

    \(=\sqrt{\left(\sqrt{3}+1\right)^2}-2\left(\sqrt{3}-1\right)=\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)\)

      \(=\sqrt{3}+1-2\sqrt{3}+2=3-\sqrt{3}=\sqrt{3}\left(\sqrt{3}-1\right)\)

g/ \(\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-2\sqrt{3}+2007\)

   \(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-2\sqrt{3}+2007\)

     \(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}-2\sqrt{3}+2007\)

       \(=2007\)

Lê Quốc Anh
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Tdq_S.Coups
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svtkvtm
31 tháng 8 2019 lúc 14:32

\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\frac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\frac{6+2\sqrt{5}}{4}=\frac{32-6-2\sqrt{5}}{4}=\frac{26-2\sqrt{5}}{4}=\frac{14-\sqrt{5}}{2}\) \(\left(\frac{9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2-\left(\frac{9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2=\left(\frac{9-2\sqrt{14}-9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)\left(\frac{9-2\sqrt{14}+9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)=\frac{-72\sqrt{14}}{\sqrt{7}-\sqrt{2}}\)

Mặc tử han
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Chiharu
29 tháng 9 2019 lúc 11:16

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Chiharu
29 tháng 9 2019 lúc 11:42

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Chiharu
29 tháng 9 2019 lúc 11:17

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