Tính \(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}+\sqrt{2+....}}}}\)
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Tính:
\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
Sửa đề: \(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
Ta có: \(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{4-2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{4-2-\sqrt{3}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2-\sqrt{3}}\)
=1
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3 tháng 8 2023 lúc 8:37
Tính
\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt[]{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2^2-\left(2+\sqrt{2+\sqrt{3}}\right)^2}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{4-2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{4-2-\sqrt{3}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2-\sqrt{3}}=\sqrt{4-3}=1\)
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23 tháng 5 2022 lúc 19:20
Tính :\(R=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
23 tháng 5 2022 lúc 21:12
Tính \(R=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}\sqrt{2-\sqrt{3}}=1\)
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24 tháng 5 2022 lúc 21:04
Tính :
\(R= \sqrt{2+\sqrt{3}}. \sqrt{2+\sqrt{2 +\sqrt{3}}}. \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
e có phát hiện mới:v cj chung lớp vs cj kia đúng hemm:v
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\(R=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\left(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\right)\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{\left(2+\sqrt{2+\sqrt{2+\sqrt{3}}}\right)\left(2-\sqrt{2+\sqrt{2+\sqrt{3}}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2^2-\left(\sqrt{2+\sqrt{2+\sqrt{3}}}\right)^2}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-\left(2+\sqrt{2+\sqrt{3}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}.\left(\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\right)\\ =\sqrt{2+\sqrt{3}}.\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\left(2-\sqrt{2+\sqrt{3}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2^2-\left(\sqrt{2+\sqrt{3}}\right)^2}\\ =\sqrt{2+\sqrt{3}}.\sqrt{4-\left(2+\sqrt{3}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}\\ =\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\\ =\sqrt{4-\sqrt{3^2}}\\ =\sqrt{4-3}\\ =\sqrt{1}\\ =1\)
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24 tháng 5 2022 lúc 19:27
Tính :\(R=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\sqrt{4-2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{4-2-\sqrt{3}}\)
=1
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=√2+√3⋅√2+√2+√3⋅√22−(2+√2+√3)2=2+3⋅2+2+3⋅22−(2+2+3)2
=√2+√3⋅√2+√2+√3⋅√4−2−√2+√3=2+3⋅2+2+3⋅4−2−2+3
=√2+√3⋅√2+√2+√3⋅√2−√2+√3=2+3⋅2+2+3⋅2−2+3
=√2+√3⋅√4−2−√3=2+3⋅4−2−3
=√2+√3⋅√2−√3=√4−3=1
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5 tháng 11 2023 lúc 17:00
Tính:( Các bạn bấm máy tính nhé)
1.\(\left(\sqrt{10}-\sqrt{14}\right).\left(\sqrt{6+\sqrt{35}}\right)\)
2. \(\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2}+\sqrt{2}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
1: \(\left(\sqrt{10}-\sqrt{14}\right)\cdot\sqrt{6+\sqrt{35}}\)
\(=\left(\sqrt{5}-\sqrt{7}\right)\cdot\sqrt{12+2\sqrt{35}}\)
\(=\left(\sqrt{5}-\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)\)
=5-7=-2
2: Sửa đề: \(\sqrt{4+\sqrt{8}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{2^2-\left(2+\sqrt{2}\right)}\)
\(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{4-2-\sqrt{2}}\)
\(=\sqrt{2\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}=\sqrt{2}\)
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tính: R=\(\sqrt{2+\sqrt{3}}+\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2\sqrt{2+\sqrt{2+\sqrt{3}}}}+\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
TÍnh \(\sqrt{2+\sqrt{3}}+\sqrt{2+\sqrt{2+\sqrt{3}}}+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}+\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
Sửa lại!
\(A=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}..\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-2-\sqrt{2+\sqrt{3}}}.\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}.\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{4-2-\sqrt{3}}=\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}=\sqrt{4-3}=1.\)
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Xem thêm câu trả lời
22 tháng 3 2022 lúc 8:36
Thực hiện các phép tính :
1. \(A=\sqrt{2-\sqrt{3}}\sqrt{2+\sqrt{2-\sqrt{3}}}\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{3}}}}\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{3}}}}}\)
2. \(B=\left(\dfrac{1}{1+\sqrt{2}}+\dfrac{2}{2+\sqrt{3}}+...+\dfrac{1}{20+\sqrt{21}}\right)\cdot2022\)
Giải chi tiết giúp mình ạ
1:
\(A=\sqrt{2-\sqrt{3}}\cdot\sqrt{2+\sqrt{2-\sqrt{3}}}\cdot\sqrt{2^2-\left(2+\sqrt{2-\sqrt{3}}\right)}\)
\(=\sqrt{2-\sqrt{3}}\cdot\sqrt{2+\sqrt{2-\sqrt{3}}}\cdot\sqrt{2-\sqrt{2-\sqrt{3}}}\)
\(=\sqrt{2-\sqrt{3}}\cdot\sqrt{4-2+\sqrt{3}}\)
\(=\sqrt{2-\sqrt{3}}\cdot\sqrt{2+\sqrt{3}}=1\)
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