\(\sqrt{2+\sqrt{ }3}.\sqrt{2+\sqrt{ }2+\sqrt{ }3}.\sqrt{2+\sqrt{ }2+\sqrt{ }2+\sqrt{ }3}.\sqrt{2+\sqrt{ }2+\sqrt{ }2+}\sqrt{\sqrt{ }2\sqrt{ }3}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
27 tháng 6 2021
a, Ta có : \(\left\{{}\begin{matrix}\sqrt{3+2\sqrt{2}}=\sqrt{2+2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\\\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\end{matrix}\right.\)
- Thay lần lượt vào A ta được :
\(A=\left(\sqrt{2}+1-\sqrt{2}+1\right)\left(\sqrt{2}-1+\sqrt{2}+1\right)=2.2\sqrt{2}=4\sqrt{2}\)
b, \(B=\sqrt{2+\sqrt{3}}\sqrt{2^2-\left(\sqrt{2+\sqrt{3}}\right)^2}=\sqrt{2+\sqrt{3}}\sqrt{4-2-\sqrt{3}}\)
\(=\sqrt{2-\sqrt{3}}\sqrt{2+\sqrt{3}}=\sqrt{4-3}=\sqrt{1}=1\)
c, \(C=\dfrac{\left(2+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)+\left(2-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}{\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}\)
\(=\dfrac{2\sqrt{2}+\sqrt{6}-2\sqrt{2-\sqrt{3}}-\sqrt{3}\sqrt{2-\sqrt{3}}+2\sqrt{2}-\sqrt{6}+2\sqrt{2+\sqrt{3}}-\sqrt{3}\sqrt{2+\sqrt{3}}}{\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}\)
\(=\dfrac{4\sqrt{2}-2\sqrt{3}\sqrt{2-\sqrt{3}}}{\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}\)
27 tháng 6 2021
a) Ta có: \(A=\left(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\right)\left(\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}\right)\)
\(=\left(\sqrt{2}+1-\sqrt{2}+1\right)\left(\sqrt{2}-1+\sqrt{2}+1\right)\)
\(=2\cdot2\sqrt{2}=4\sqrt{2}\)
7 tháng 10 2022
a: \(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{4-2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{4-2-\sqrt{3}}\)
\(=\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=1\)
b: \(=\sqrt{2+\sqrt{2}}\cdot\sqrt{3+\sqrt{7+\sqrt{2}}}\cdot\sqrt{9-6-\sqrt{7+\sqrt{2}}}\)
\(=\sqrt{2+\sqrt{2}}\cdot\sqrt{9-7-\sqrt{2}}\)
\(=\sqrt{2}\)
3 tháng 7 2021
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
3 tháng 8 2023
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2^2-\left(2+\sqrt{2+\sqrt{3}}\right)^2}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{4-2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{4-2-\sqrt{3}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2-\sqrt{3}}=\sqrt{4-3}=1\)
VD
25 tháng 5 2022
\(R=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\left(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\right)\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{\left(2+\sqrt{2+\sqrt{2+\sqrt{3}}}\right)\left(2-\sqrt{2+\sqrt{2+\sqrt{3}}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2^2-\left(\sqrt{2+\sqrt{2+\sqrt{3}}}\right)^2}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-\left(2+\sqrt{2+\sqrt{3}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}.\left(\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\right)\\ =\sqrt{2+\sqrt{3}}.\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\left(2-\sqrt{2+\sqrt{3}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2^2-\left(\sqrt{2+\sqrt{3}}\right)^2}\\ =\sqrt{2+\sqrt{3}}.\sqrt{4-\left(2+\sqrt{3}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}\\ =\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\\ =\sqrt{4-\sqrt{3^2}}\\ =\sqrt{4-3}\\ =\sqrt{1}\\ =1\)
13 tháng 6 2023
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\sqrt{4-2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{4-2-\sqrt{3}}\)
=1
3 tháng 8 2023
=√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
13 tháng 6 2023
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}\sqrt{2-\sqrt{3}}=1\)
24 tháng 5 2022
e có phát hiện mới:v cj chung lớp vs cj kia đúng hemm:v
VD
25 tháng 5 2022
\(R=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\left(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\right)\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{\left(2+\sqrt{2+\sqrt{2+\sqrt{3}}}\right)\left(2-\sqrt{2+\sqrt{2+\sqrt{3}}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2^2-\left(\sqrt{2+\sqrt{2+\sqrt{3}}}\right)^2}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-\left(2+\sqrt{2+\sqrt{3}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}.\left(\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\right)\\ =\sqrt{2+\sqrt{3}}.\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\left(2-\sqrt{2+\sqrt{3}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2^2-\left(\sqrt{2+\sqrt{3}}\right)^2}\\ =\sqrt{2+\sqrt{3}}.\sqrt{4-\left(2+\sqrt{3}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}\\ =\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\\ =\sqrt{4-\sqrt{3^2}}\\ =\sqrt{4-3}\\ =\sqrt{1}\\ =1\)