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VƯƠN CAO VIỆT NAM
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Akai Haruma
23 tháng 8 2019 lúc 9:17

Lời giải:

\(M=\frac{2\sqrt{4-\sqrt{5+\sqrt{20+1+2\sqrt{20.1}}}}}{\sqrt{10}-\sqrt{2}}=\frac{2\sqrt{4-\sqrt{5+\sqrt{(\sqrt{20}+1)^2}}}}{\sqrt{10}-\sqrt{2}}\)

\(=\frac{2\sqrt{4-\sqrt{5+\sqrt{20}+1}}}{\sqrt{10}-\sqrt{2}}=\frac{2\sqrt{4-\sqrt{5+1+2\sqrt{5}}}}{\sqrt{10}-\sqrt{2}}\)

\(=\frac{2\sqrt{4-\sqrt{(\sqrt{5}+1)^2}}}{\sqrt{10}-\sqrt{2}}=\frac{2\sqrt{4-(\sqrt{5}+1)}}{\sqrt{2}(\sqrt{5}-1)}=\frac{\sqrt{2}.\sqrt{3-\sqrt{5}}}{\sqrt{5}-1}\)

\(=\frac{\sqrt{6-2\sqrt{5}}}{\sqrt{5}-1}=\frac{\sqrt{5+1-2\sqrt{5}}}{\sqrt{5}-1}=\frac{\sqrt{(\sqrt{5}-1)^2}}{\sqrt{5}-1}=\frac{\sqrt{5}-1}{\sqrt{5}-1}=1\)

Akai Haruma
17 tháng 8 2019 lúc 0:06

Lời giải:

\(M=\frac{2\sqrt{4-\sqrt{5+\sqrt{20+1+2\sqrt{20.1}}}}}{\sqrt{10}-\sqrt{2}}=\frac{2\sqrt{4-\sqrt{5+\sqrt{(\sqrt{20}+1)^2}}}}{\sqrt{10}-\sqrt{2}}\)

\(=\frac{2\sqrt{4-\sqrt{5+\sqrt{20}+1}}}{\sqrt{10}-\sqrt{2}}=\frac{2\sqrt{4-\sqrt{5+1+2\sqrt{5}}}}{\sqrt{10}-\sqrt{2}}\)

\(=\frac{2\sqrt{4-\sqrt{(\sqrt{5}+1)^2}}}{\sqrt{10}-\sqrt{2}}=\frac{2\sqrt{4-(\sqrt{5}+1)}}{\sqrt{2}(\sqrt{5}-1)}=\frac{\sqrt{2}.\sqrt{3-\sqrt{5}}}{\sqrt{5}-1}\)

\(=\frac{\sqrt{6-2\sqrt{5}}}{\sqrt{5}-1}=\frac{\sqrt{5+1-2\sqrt{5}}}{\sqrt{5}-1}=\frac{\sqrt{(\sqrt{5}-1)^2}}{\sqrt{5}-1}=\frac{\sqrt{5}-1}{\sqrt{5}-1}=1\)

Soldier Winter
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Hoàng Kiệt
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Nguyễn Thị Ngọc Diệp
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trần thị trâm anh
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Thúy Nga
10 tháng 9 2018 lúc 13:16

\(A=\dfrac{4\sqrt{4-\sqrt{5+\sqrt{21+\sqrt{80}}}}}{\sqrt{10}-\sqrt{2}}=\dfrac{4\sqrt{4-\sqrt{5+\sqrt{\left(\sqrt{20}+1\right)^2}}}}{\sqrt{10}-\sqrt{2}}\\ =\dfrac{4\sqrt{4-\sqrt{\left(\sqrt{5}+1\right)^2}}}{\sqrt{10}-\sqrt{2}}=\dfrac{4\sqrt{3-\sqrt{5}}}{\sqrt{10}-\sqrt{2}}\\ =\dfrac{2\sqrt{\left(\sqrt{10}-\sqrt{2}\right)^2}}{\sqrt{10}-\sqrt{2}}=\dfrac{2\left(\sqrt{10}-\sqrt{2}\right)}{\sqrt{10}-\sqrt{2}}\\ =2\)

Phan Na 1802
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nguyễn thị thanh thùy
20 tháng 10 2018 lúc 21:12

 a) \(\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{7\left(\sqrt{3}+\sqrt{5}\right)}}=\) \(\frac{\sqrt{2}}{\sqrt{7}}\)

 b ) \(\frac{15\sqrt{2}+9\sqrt{3}}{3\sqrt{3}+3\sqrt{5}}=\frac{3\left(5\sqrt{2}+3\sqrt{3}\right)}{3\left(\sqrt{3}+\sqrt{5}\right)}\)\(=\frac{5\sqrt{2}+3\sqrt{3}}{\sqrt{3}+\sqrt{5}}\)

c)\(\frac{\sqrt{2}-\sqrt{6}+\sqrt{3}-\sqrt{9}+\sqrt{4}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\) =  \(\frac{\sqrt{2}\left(1-\sqrt{3}\right)+\sqrt{3}\left(1-\sqrt{3}\right)+\sqrt{4}\left(1-\sqrt{3}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)\(=\frac{\left(1-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1-\sqrt{3}\)

 d) \(\frac{\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{5}-1}=\frac{\sqrt{5}-1}{\sqrt{5}-1}=1\)

poppy Trang
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tran nguyen bao quan
22 tháng 9 2018 lúc 14:18

\(A=\dfrac{2\sqrt{4-\sqrt{5+\sqrt{21+\sqrt{80}}}}}{\sqrt{10}-\sqrt{2}}=\dfrac{2\sqrt{4-\sqrt{5+\sqrt{20+4\sqrt{5}+1}}}}{\sqrt{10}-\sqrt{2}}\)\(=\dfrac{2\sqrt{4-\sqrt{5+\sqrt{\left(2\sqrt{5}+1\right)^2}}}}{\sqrt{10}-\sqrt{2}}=\dfrac{2\sqrt{4-\sqrt{5+\left(2\sqrt{5}+1\right)}}}{\sqrt{10}-\sqrt{2}}=\)\(=\dfrac{2\sqrt{4-\sqrt{\left(\sqrt{5}+1\right)^2}}}{\sqrt{10}-\sqrt{2}}=\dfrac{2\sqrt{4-\sqrt{5}-1}}{\sqrt{10}-\sqrt{2}}=\dfrac{\sqrt{2}.\sqrt{6-2\sqrt{5}}}{\sqrt{10}-\sqrt{2}}=\dfrac{\sqrt{2}.\sqrt{5-2\sqrt{5}+1}}{\sqrt{10}-\sqrt{2}}=\dfrac{\sqrt{2}.\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{10}-\sqrt{2}}=\dfrac{\sqrt{2}.\left(\sqrt{5}-1\right)}{\sqrt{10}-\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{2}}{\sqrt{10}-\sqrt{2}}=1\)

hoàng hà diệp
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Nguyễn Xuân Anh
2 tháng 10 2018 lúc 22:20

\(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}.\)

\(\Rightarrow A^2=4+\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{2}}\right)\left(4-\sqrt{10+2\sqrt{2}}\right)}+4-\sqrt{10+2\sqrt{5}}\)

          \(=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)

          \(=8+2\sqrt{6-2\sqrt{5}}\)

          \(=8+2\sqrt{5-2\sqrt{5.1}+1}=8+2\left(\sqrt{5}-1\right)\)

           \(=8+2\sqrt{5}-2=6+2\sqrt{5}\)

          \(=\left(\sqrt{5}+1\right)^2\)

\(\Rightarrow A=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)

\(B=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+...+\frac{1}{\sqrt{2001}+\sqrt{2005}}\)

    \(=\frac{1-\sqrt{5}}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}+\frac{\sqrt{5}-\sqrt{9}}{\left(\sqrt{5}+\sqrt{9}\right)\left(\sqrt{5}-\sqrt{9}\right)}+...+\frac{\sqrt{2001}-\sqrt{2005}}{\left(\sqrt{2001}+\sqrt{2005}\right)\left(\sqrt{2001}-\sqrt{2005}\right)}\)

\(=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)

\(=-\frac{1}{4}\left(1-\sqrt{5}+\sqrt{5}-\sqrt{9}+....+\sqrt{2001}-\sqrt{2005}\right)\)

\(=-\frac{1}{4}\left(1-\sqrt{2005}\right)\)

\(=10,94430659\)

\(\text{Lm hơi vắn tắt thông cảm nha!!}\)

AK-47
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HT.Phong (9A5)
17 tháng 7 2023 lúc 8:13

1) \(\sqrt{6+4\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)

\(=\sqrt{2^2+2\cdot2\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{3^2-2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}\)

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

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

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

\(=2\sqrt{2}-1\)

2) \(\sqrt{21-4\sqrt{5}}+\sqrt{21+4\sqrt{5}}\)

\(=\sqrt{20-4\sqrt{5}+1}+\sqrt{20+4\sqrt{5}+1}\)

\(=\sqrt{\left(2\sqrt{5}\right)^2-2\sqrt{5}\cdot2\cdot1+1^2}+\sqrt{\left(2\sqrt{5}\right)^2+2\sqrt{5}\cdot2\cdot1-1^2}\)

\(=\sqrt{\left(2\sqrt{5}-1\right)^2}+\sqrt{\left(2\sqrt{5}+1\right)^2}\)

\(=\left|2\sqrt{5}-1\right|+\left|2\sqrt{5}+1\right|\)

\(=2\sqrt{5}-1+2\sqrt{5}+1\)

\(=4\sqrt{5}\)