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Nhóc vậy
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Nguyễn Khánh Ly
9 tháng 12 2017 lúc 19:21

\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\) =\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-12\sqrt{5}+9}}}\)=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)=\(\sqrt{\sqrt{5}-\sqrt{3-\left|2\sqrt{5}-3\right|}}\)=\(\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)=\(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)=\(\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}\)=\(\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)=\(\sqrt{\sqrt{5}-\left|\sqrt{5}-1\right|}\)=\(\sqrt{\sqrt{5}-\sqrt{5}+1}\)=\(\sqrt{1}\)=1( là số nguyên )

=> Số đã cho nguyên

Nguyễn Hoàng trung
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Lê Thị Thục Hiền
25 tháng 6 2021 lúc 13:04

\(Q=\sqrt{\sqrt{5}-1}\left(\sqrt{8-\sqrt{5}+2\sqrt{5\sqrt{5}-3}}-\sqrt{7-\sqrt{20}}\right)\)

\(\Rightarrow\)\(Q^2=\left(\sqrt{5}-1\right)\left(8-\sqrt{5}+2\sqrt{5\sqrt{5}-3}+7-\sqrt{20}-2\sqrt{\left(7-\sqrt{20}\right)\left(8-\sqrt{5}+2\sqrt{5\sqrt{5}-3}\right)}\right)\)

\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\sqrt{\left(7-2\sqrt{5}\right)\left(8-\sqrt{5}\right)+2\left(7-2\sqrt{5}\right)\sqrt{5\sqrt{5}-3}}\right)\)

\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\sqrt{66-23\sqrt{5}+2\left(7-2\sqrt{5}\right)\sqrt{5\sqrt{5}-3}}\right)\)

\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\sqrt{\left(49-28\sqrt{5}+20\right)+2\left(7-2\sqrt{5}\right)\sqrt{5\sqrt{5}-3}+\left(5\sqrt{5}-3\right)}\right)\)

\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\sqrt{\left(7-2\sqrt{5}\right)^2+2\left(7-2\sqrt{5}\right)\sqrt{5\sqrt{5}-3}+\left(5\sqrt{5}-3\right)}\right)\)

\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\sqrt{\left(7-2\sqrt{5}+\sqrt{5\sqrt{5}-3}\right)^2}\right)\)

\(=\left(\sqrt{5}-1\right)\left(15-3\sqrt{5}+2\sqrt{5\sqrt{5}-3}-2\left(7-2\sqrt{5}+\sqrt{5\sqrt{5}-3}\right)\right)\)

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

\(\Rightarrow Q^2=4\) \(\Rightarrow Q\) nguyên 

Uyên Dii
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Lê Đình Thái
21 tháng 9 2017 lúc 19:38

a) A=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

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

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

b) B=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)\)

=\(\left(4\sqrt{10}+\sqrt{150}-4\sqrt{6}-\sqrt{90}\right)\left(\sqrt{4-\sqrt{15}}\right)\)

=\(\left(4\sqrt{10}+5\sqrt{6}-4\sqrt{6}-3\sqrt{10}\right)\left(\sqrt{4-\sqrt{15}}\right)\)

=\(\left(\sqrt{10}+\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)

=\(5-\sqrt{15}+\sqrt{15}-3=2\)(đpcm)

shizami
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Nguyễn Lê Phước Thịnh
11 tháng 12 2021 lúc 21:29

1: Thay x=16 vào A, ta được:

\(A=\dfrac{6-2\cdot4}{4-5}=\dfrac{-2}{-1}=2\)

Ngô Hoài Thanh
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Trần Lê Quỳnh Anh
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Charlet
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Ngô Hoài Thanh
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Rộp Rộp Rộp
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Kiyotaka Ayanokoji
27 tháng 7 2020 lúc 10:00

Trả lời:

\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-12\sqrt{5}+9}}}\)

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

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

\(A=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)

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

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

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

\(A=\sqrt{1}\)

\(A=1\)

\(B=\frac{\left(5+2\sqrt{6}\right).\left(49-20\sqrt{6}\right).\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)

\(B=\frac{\left(3+2\sqrt{6}+2\right).\left(49-20\sqrt{6}\right).\sqrt{3-2\sqrt{6}+2}}{9\sqrt{3}-11\sqrt{2}}\)

\(B=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2.\left(49-20\sqrt{6}\right).\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{9\sqrt{3}-11\sqrt{2}}\)

\(B=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2.\left(49-20\sqrt{6}\right).\left(\sqrt{3}-\sqrt{2}\right)}{9\sqrt{33}-11\sqrt{2}}\)

\(B=\frac{\left(\sqrt{3}+\sqrt{2}\right).\left(\sqrt{3}-\sqrt{2}\right).\left(\sqrt{3}+\sqrt{2}\right).\left(49-20\sqrt{6}\right)}{9\sqrt{3}-11\sqrt{2}}\)

\(B=\frac{\left(3-2\right).\left(49\sqrt{3}-60\sqrt{2}+49\sqrt{2}-40\sqrt{3}\right)}{9\sqrt{3}-11\sqrt{2}}\)

\(B=\frac{1.\left(9\sqrt{3}-11\sqrt{2}\right)}{9\sqrt{3}-11\sqrt{2}}\)

\(B=1\)

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Nobi Nobita
20 tháng 9 2020 lúc 16:01

a) Ta có: \(\sqrt{29-12\sqrt{5}}=\sqrt{20-12\sqrt{5}+9}=\sqrt{\left(2\sqrt{5}-3\right)^2}\)

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

\(\Rightarrow\sqrt{3-\sqrt{29-12\sqrt{5}}}=\sqrt{3-\left(2\sqrt{5}-3\right)}=\sqrt{3-2\sqrt{5}+3}\)

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

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

\(\Leftrightarrow A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)

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

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