PT
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
TT
1
NC
30 tháng 8 2020
Bài làm:
Ta có: \(\sqrt{4+\sqrt{5\sqrt{3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3+5\sqrt{28-10\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3+5\sqrt{\left(5-\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3+5\left(5-\sqrt{3}\right)}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{28-5\sqrt{3}}}}\)
Đến đây chịu-.-
H
2 tháng 6 2018
\(\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{10}+\sqrt{3+\sqrt{5}}\right)}-\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{10}+\sqrt{3-\sqrt{5}}\right)}\)
=\(\dfrac{3\sqrt{2}+\sqrt{10}}{\sqrt{20}+\sqrt{6+2\sqrt{5}}}-\dfrac{3\sqrt{2}-\sqrt{10}}{\sqrt{20}+\sqrt{6-2\sqrt{5}}}\)
=\(\dfrac{3\sqrt{2}+\sqrt{10}}{\sqrt{20}+\sqrt{5+2\sqrt{5}+1}}-\dfrac{3\sqrt{2}-\sqrt{10}}{\sqrt{20}+\sqrt{5-2\sqrt{5}+1}}\)
=\(\dfrac{3\sqrt{2}+\sqrt{10}}{\sqrt{20}+\sqrt{\left(\sqrt{5}+1\right)^2}}-\dfrac{3\sqrt{2}-\sqrt{10}}{\sqrt{20}+\sqrt{\left(\sqrt{5}-1\right)^2}}\)
=\(\dfrac{3\sqrt{2}+\sqrt{10}}{\sqrt{20}+\sqrt{5}+1}-\dfrac{3\sqrt{2}-\sqrt{10}}{\sqrt{20}+\sqrt{5}-1}\)
=\(\dfrac{3\sqrt{2}+\sqrt{10}}{3\sqrt{5}+1}-\dfrac{3\sqrt{2}-\sqrt{10}}{3\sqrt{5}-1}\)
=\(\dfrac{(3\sqrt{2}+\sqrt{10})\left(3\sqrt{5}-1\right)}{(3\sqrt{5}+1)\left(3\sqrt{5}-1\right)}-\dfrac{(3\sqrt{2}-\sqrt{10})\left(3\sqrt{5}+1\right)}{(3\sqrt{5}+1)\left(3\sqrt{5}-1\right)}\)
=\(\dfrac{9\sqrt{10}-3\sqrt{2}+3\sqrt{50}-\sqrt{10}-\left(9\sqrt{10}+3\sqrt{2}-3\sqrt{50}-\sqrt{10}\right)}{(3\sqrt{5}+1)\left(3\sqrt{5}-1\right)}\)
=\(\dfrac{-6\sqrt{2}+6\sqrt{50}}{\left(3\sqrt{5}\right)^2-1}\)=\(\dfrac{-6\left(\sqrt{2}-\sqrt{50}\right)}{45-1}\)
=\(\dfrac{-6\sqrt{2}\left(1-5\right)}{44}=\dfrac{24\sqrt{2}}{44}=\dfrac{6\sqrt{2}}{11}\)
BG
2
14 tháng 7 2021
Ta có: \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)
\(=\sqrt{25}=5\)
14 tháng 7 2021
\(\sqrt{5\sqrt{3+5\sqrt{48}-10\sqrt{7+4\sqrt{3}}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{48}-10\sqrt{4+2.2\sqrt{3+\left(\sqrt{3}\right)^2}}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{48-10.\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{48-10.\sqrt{\left(2+\sqrt{3}\right)}}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{48-20+10}\sqrt{3}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{28+10}\sqrt{3}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{\left(\sqrt{3}\right)^2}+2.5.\sqrt{3}+5^2}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{\left(\sqrt{3}+5\right)^2}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{\left(\sqrt{3}+5\right)}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{3+10}}}\)
=\(10\sqrt{3+10}\)
=\(\sqrt{10\left(\sqrt{3+1}\right)}\)
PA
12 tháng 7 2017
\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+\sqrt{48}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2-\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-20+10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
= 5
\(\dfrac{\sqrt{3}-\sqrt{5+\sqrt{24}}+\sqrt{\sqrt{72}+11}}{\sqrt{6+\sqrt{20}}+\sqrt{2}-\sqrt{7+\sqrt{40}}}\)
\(=\dfrac{\sqrt{3}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{2}\right)^2}}{\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{2}-\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}}\)
\(=\dfrac{\sqrt{3}-\sqrt{2}-\sqrt{3}+3+\sqrt{2}}{\sqrt{5}+1+\sqrt{2}-\sqrt{2}-\sqrt{5}}\)
\(=3\)
Đáp án là 0,301823955......
một câu vô nghĩa