K
Khách

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

19 tháng 6 2017

a) \(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\dfrac{\left(245-100\sqrt{6}+98\sqrt{6}-240\right)\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\dfrac{\left(5-2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)}{9\sqrt{3}-11\sqrt{2}}\)

\(=\dfrac{5\sqrt{3}-5\sqrt{2}-2\sqrt{18}+2\sqrt{12}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\dfrac{5\sqrt{3}-5\sqrt{2}-6\sqrt{2}+4\sqrt{3}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\dfrac{9\sqrt{3}-11\sqrt{2}}{9\sqrt{3}-11\sqrt{2}}\)

\(=1\)

19 tháng 6 2017

b)

\(\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)

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

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

\(=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{3\sqrt{3\left(2+\sqrt{3}\right)}-2\sqrt{18}+3\sqrt{2+\sqrt{3}}}{6\sqrt{3}}}\)

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

\(=\dfrac{3\sqrt{\left(2+\sqrt{3}\right)\cdot3}}{3\sqrt{6+3\sqrt{3}}-6\sqrt{2}+3\sqrt{2+\sqrt{3}}}\)

\(=\dfrac{3\sqrt{\left(2+\sqrt{3}\right)\cdot3}}{3\left(\sqrt{6+3\sqrt{3}}-2\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}\)

\(=\dfrac{\sqrt{\left(2+\sqrt{3}\right)\cdot3}}{\sqrt{6+3\sqrt{3}}-2\sqrt{2}+\sqrt{2+\sqrt{3}}}\)

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

\(=\dfrac{\sqrt{\left(6+3\sqrt{3}\right)\left(-\sqrt{3}+2+\sqrt{3}\right)}}{-2\sqrt{3}}\)

\(=\dfrac{\sqrt{\left(6+3\sqrt{3}\right)\cdot2}}{-2\sqrt{3}}\)

\(=\dfrac{\sqrt{12+6\sqrt{3}}}{-2\sqrt{3}}\)

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

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

\(=-\dfrac{\left(3+\sqrt{3}\right)\sqrt{3}}{6}\)

\(=-\dfrac{3\sqrt{3}+3}{6}\)

\(=-\dfrac{3\left(\sqrt{3}+3\right)}{6}\)

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

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

b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)

\(=-\sqrt{3}\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)

\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)

c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)

10 tháng 7 2017

\(A=\sqrt{8}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =2\sqrt{2}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =4\sqrt{2}+4\sqrt{7}\)

10 tháng 7 2017

\(B=\left(3+2\sqrt{6}+2\right)\left(25-20\sqrt{6}+24\right)\sqrt{3-2\sqrt{6}+2}\\ =\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}-\sqrt{2}\right)\\ =\left(\sqrt{3}+\sqrt{2}\right)\left(3-2\sqrt{6}+2\right)^2\\ =\left(\sqrt{3}-\sqrt{2}\right)^3\\ =9\sqrt{3}-11\sqrt{2}\)

a: Ta có: \(A=\left(\dfrac{6+\sqrt{20}}{3+\sqrt{5}}+\dfrac{\sqrt{14}-\sqrt{2}}{\sqrt{7}-1}\right):\left(2+\sqrt{2}\right)\)

\(=\left(2+\sqrt{2}\right):\left(2+\sqrt{2}\right)\)

=1

b: Ta có: \(B=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-\dfrac{11}{2\sqrt{3}+1}\)

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

=1

24 tháng 8 2021

a)\(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)

\(=2.\sqrt{\dfrac{4^2}{3}}-3.\sqrt{\dfrac{1}{3.3^2}}-6\sqrt{\dfrac{2^2}{3.5^2}}\)

\(=2.\dfrac{4}{\sqrt{3}}-3.\dfrac{1}{3\sqrt{3}}-6.\dfrac{2}{5\sqrt{3}}=\dfrac{8}{\sqrt{3}}-\dfrac{1}{\sqrt{3}}-\dfrac{12}{5\sqrt{3}}\)\(=\dfrac{23}{5\sqrt{3}}=\dfrac{23\sqrt{3}}{15}\)

b)\(\left(6\sqrt{\dfrac{8}{9}}-5\sqrt{\dfrac{32}{25}}+14\sqrt{\dfrac{18}{49}}\right).\sqrt{\dfrac{1}{2}}\)

\(=6\sqrt{\dfrac{8}{9}.\dfrac{1}{2}}-5\sqrt{\dfrac{32}{25}.\dfrac{1}{2}}+14\sqrt{\dfrac{18}{49}.\dfrac{1}{2}}\)

\(=6\sqrt{\dfrac{4}{9}}-5\sqrt{\dfrac{16}{25}}+14\sqrt{\dfrac{9}{49}}\)\(=6.\dfrac{2}{3}-5.\dfrac{4}{5}+14.\dfrac{3}{7}=6\)

c)\(\sqrt{\left(\sqrt{2}-2\right)^2}-\sqrt{6+4\sqrt{2}}=\left|\sqrt{2}-2\right|-\sqrt{4+2.2\sqrt{2}+2}=2-\sqrt{2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)

\(=2-\sqrt{2}-\left(2+\sqrt{2}\right)=-2\sqrt{2}\)

24 tháng 11 2021

\(a,=6\sqrt{2}-3-6\sqrt{2}=-3\\ b,=12\sqrt{3}-2\sqrt{5}-6\sqrt{3}+5\sqrt{5}=6\sqrt{3}+3\sqrt{5}\\ c,=\sqrt{3}-1-\sqrt{3}=-1\\ d,=\sqrt{6}-\dfrac{5\left(\sqrt{6}+1\right)}{5}=\sqrt{6}-\sqrt{6}-1=-1\)