K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

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-.-

2 tháng 6 2018

\(\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{10}+\sqrt{3+\sqrt{5}}\right)}-\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{10}+\sqrt{3-\sqrt{5}}\right)}\)

=\(\dfrac{3\sqrt{2}+\sqrt{10}}{\sqrt{20}+\sqrt{6+2\sqrt{5}}}-\dfrac{3\sqrt{2}-\sqrt{10}}{\sqrt{20}+\sqrt{6-2\sqrt{5}}}\)

=\(\dfrac{3\sqrt{2}+\sqrt{10}}{\sqrt{20}+\sqrt{5+2\sqrt{5}+1}}-\dfrac{3\sqrt{2}-\sqrt{10}}{\sqrt{20}+\sqrt{5-2\sqrt{5}+1}}\)

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

=\(\dfrac{3\sqrt{2}+\sqrt{10}}{\sqrt{20}+\sqrt{5}+1}-\dfrac{3\sqrt{2}-\sqrt{10}}{\sqrt{20}+\sqrt{5}-1}\)

=\(\dfrac{3\sqrt{2}+\sqrt{10}}{3\sqrt{5}+1}-\dfrac{3\sqrt{2}-\sqrt{10}}{3\sqrt{5}-1}\)

=\(\dfrac{(3\sqrt{2}+\sqrt{10})\left(3\sqrt{5}-1\right)}{(3\sqrt{5}+1)\left(3\sqrt{5}-1\right)}-\dfrac{(3\sqrt{2}-\sqrt{10})\left(3\sqrt{5}+1\right)}{(3\sqrt{5}+1)\left(3\sqrt{5}-1\right)}\)

=\(\dfrac{9\sqrt{10}-3\sqrt{2}+3\sqrt{50}-\sqrt{10}-\left(9\sqrt{10}+3\sqrt{2}-3\sqrt{50}-\sqrt{10}\right)}{(3\sqrt{5}+1)\left(3\sqrt{5}-1\right)}\)

=\(\dfrac{-6\sqrt{2}+6\sqrt{50}}{\left(3\sqrt{5}\right)^2-1}\)=\(\dfrac{-6\left(\sqrt{2}-\sqrt{50}\right)}{45-1}\)

=\(\dfrac{-6\sqrt{2}\left(1-5\right)}{44}=\dfrac{24\sqrt{2}}{44}=\dfrac{6\sqrt{2}}{11}\)

Ta có: \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)

\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}\)

\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)

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

\(=\sqrt{25}=5\)

\(\sqrt{5\sqrt{3+5\sqrt{48}-10\sqrt{7+4\sqrt{3}}}}\)

=\(\sqrt{5\sqrt{3+5\sqrt{48}-10\sqrt{4+2.2\sqrt{3+\left(\sqrt{3}\right)^2}}}}\)

=\(\sqrt{5\sqrt{3+5\sqrt{48-10.\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)

=\(\sqrt{5\sqrt{3+5\sqrt{48-10.\sqrt{\left(2+\sqrt{3}\right)}}}}\)

=\(\sqrt{5\sqrt{3+5\sqrt{48-20+10}\sqrt{3}}}\)

=\(\sqrt{5\sqrt{3+5\sqrt{28+10}\sqrt{3}}}\)

=\(\sqrt{5\sqrt{3+5\sqrt{\left(\sqrt{3}\right)^2}+2.5.\sqrt{3}+5^2}}\)

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

=\(\sqrt{5\sqrt{3+5\sqrt{\left(\sqrt{3}+5\right)}}}\)

=\(\sqrt{5\sqrt{3+5\sqrt{3+10}}}\)

=\(10\sqrt{3+10}\)

=\(\sqrt{10\left(\sqrt{3+1}\right)}\)

24 tháng 5 2017

Hỏi đáp Toán

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

\(\dfrac{\sqrt{3}-\sqrt{5+\sqrt{24}}+\sqrt{\sqrt{72}+11}}{\sqrt{6+\sqrt{20}}+\sqrt{2}-\sqrt{7+\sqrt{40}}}\)

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

\(=\dfrac{\sqrt{3}-\sqrt{2}-\sqrt{3}+3+\sqrt{2}}{\sqrt{5}+1+\sqrt{2}-\sqrt{2}-\sqrt{5}}\)

\(=3\)

26 tháng 7 2018

E = \(6x+\sqrt{9x^2-12x+4}\)

E = \(6x+\sqrt{\left(3x-2\right)^2}\)

E = \(6x+\left|3x-2\right|\)

E = \(6x+3x-2\)

E = \(9x-2\)

F = \(5x-\sqrt{x^2+4x+4}\)

F = \(5x-\sqrt{\left(x+2\right)^2}\)

F = \(5x-\left|x+2\right|\)

F = \(5x-x+2\)

F = \(4x+2\)