\(\dfrac{6}{\sqrt{5}+\sqrt{11}}\)-\(\dfrac{11+\sqrt{11}}{\sqrt{11}+1}\)+\(\dfrac{1}{2\sqrt{5}}\)
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
RK
23 tháng 6 2021
a) \(\sqrt{\dfrac{1}{8}}\cdot\sqrt{2}\cdot\sqrt{125}\cdot\sqrt{\dfrac{1}{5}}\) = \(\sqrt{\dfrac{1}{8}\cdot2}.\sqrt{125\cdot\dfrac{1}{5}}=\sqrt{\dfrac{1}{4}}.\sqrt{25}=\dfrac{1}{2}\cdot5=2,5\)
RK
23 tháng 6 2021
b)\(\sqrt{\sqrt{2}-1}.\sqrt{\sqrt{2}+1}=\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2-1}=1\)
PT
4 tháng 11 2018
\(=\dfrac{5.\left(\sqrt{11}+\sqrt{6}\right)}{5}-\dfrac{2.\left(\sqrt{6}+2\right)}{2}+\dfrac{\sqrt{11}.\left(\sqrt{11}-1\right)}{1-\sqrt{11}}\)
= \(\sqrt{11}+\sqrt{6}-\sqrt{6}-2-\sqrt{11}\)
= -2
4 tháng 11 2018
\(=\dfrac{5\left(\sqrt{11}+\sqrt{6}\right)}{\left(\sqrt{11}-\sqrt{6}\right)\left(\sqrt{11}+\sqrt{6}\right)}-\dfrac{2\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}+\dfrac{\sqrt{11}\left(\sqrt{11}-1\right)}{\left(\sqrt{11}-1\right)}\)
\(=\dfrac{5\left(\sqrt{11}+\sqrt{6}\right)}{5}-\dfrac{2\left(\sqrt{6}+2\right)}{2}+\dfrac{\sqrt{11}\left(\sqrt{11}-1\right)}{\left(\sqrt{11}-1\right)}\)
\(=\left(\sqrt{11}+\sqrt{6}\right)-\left(\sqrt{6}+2\right)+\sqrt{11}\)
\(=2\sqrt{11}-2\)
AH
Akai Haruma
Giáo viên
27 tháng 10 2018
\(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-....-\frac{1}{\sqrt{24}-\sqrt{25}}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{(\sqrt{1}-\sqrt{2})(\sqrt{1}+\sqrt{2})}-\frac{\sqrt{2}+\sqrt{3}}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}+\frac{\sqrt{3}+\sqrt{4}}{(\sqrt{3}-\sqrt{4})(\sqrt{3}+\sqrt{4})}-...-\frac{\sqrt{24}+\sqrt{25}}{(\sqrt{24}-\sqrt{25})(\sqrt{24}+\sqrt{25})}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{-1}-\frac{\sqrt{2}+\sqrt{3}}{-1}+\frac{\sqrt{3}+\sqrt{4}}{-1}-...-\frac{\sqrt{24}+\sqrt{25}}{-1}\)
\(=\frac{(1+\sqrt{2})-(\sqrt{2}+\sqrt{3})+(\sqrt{3}+\sqrt{4})-...-(\sqrt{24}+\sqrt{25})}{-1}\)
\(=\frac{1-\sqrt{25}}{-1}=4\)
AH
Akai Haruma
Giáo viên
27 tháng 10 2018
\(B=\frac{5}{4+\sqrt{11}}+\frac{11-3\sqrt{11}}{\sqrt{11}-3}-\frac{4}{\sqrt{5}-1}+\sqrt{(\sqrt{5}-2)^2}\)
\(=\frac{5(4-\sqrt{11})}{(4+\sqrt{11})(4-\sqrt{11})}+\frac{\sqrt{11}(\sqrt{11}-3)}{\sqrt{11}-3}-\frac{4(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}+\sqrt{5}-2\)
\(=\frac{5(4-\sqrt{11})}{5}+\sqrt{11}-\frac{4(\sqrt{5}+1)}{4}+\sqrt{5}-2\)
\(=4-\sqrt{11}+\sqrt{11}-(\sqrt{5}+1)+\sqrt{5}-2\)
\(=1\)
14 tháng 10 2022
\(VT=4+\sqrt{11}+\dfrac{3}{2}-\dfrac{1}{2}\sqrt{7}-4-2\sqrt{7}-\dfrac{1}{2}\sqrt{7}+\dfrac{5}{2}\)
\(=4+\sqrt{11}-3\sqrt{7}\)
30 tháng 9 2021
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
26 tháng 8 2021
a: Ta có: \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}-11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}-11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}-11\right)\)
\(=127-22\sqrt{6}\)
b: Ta có: \(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
\(=\left(1-\sqrt{5}\right)\left(-1-\sqrt{5}\right)\)
=-1+5
=4
26 tháng 9 2018
\(\dfrac{5\left(4+\sqrt{11}\right)}{\left(4+\sqrt{11}\right)\left(4-\sqrt{11}\right)}+\dfrac{3-\sqrt{7}}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}-\dfrac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\dfrac{\sqrt{7}-5}{2}\)\(=\dfrac{\left(4+\sqrt{11}\right)5}{16-11}+\dfrac{3-\sqrt{7}}{9-7}-\dfrac{6\left(\sqrt{7}+2\right)}{7-4}-\dfrac{\sqrt{7}-5}{2}\)
\(=4+\sqrt{11}-\dfrac{3-\sqrt{7}}{2}-2\left(\sqrt{7}+2\right)-\dfrac{\sqrt{7}-5}{2}=\dfrac{8+2\sqrt{11}-3+\sqrt{7}-4\sqrt{7}-8-\sqrt{7}+5}{2}=\dfrac{2\sqrt{11}-4\sqrt{7}+2}{2}=1+\sqrt{11}-2\sqrt{7}\)
HT
1
6 tháng 8 2018
\(\dfrac{5+7\sqrt{5}}{\sqrt{5}}+\dfrac{11+\sqrt{11}}{1+\sqrt{11}}=\dfrac{\sqrt{5}\left(\sqrt{5}+7\right)}{\sqrt{5}}+\dfrac{\sqrt{11}\left(1+\sqrt{11}\right)}{1+\sqrt{11}}=\sqrt{5}+7+\sqrt{11}\)
`6/(sqrt11+sqrt5)-(11+sqrt11)/(sqrt11+1)+1/(2sqrt5)`
`=(6(sqrt11-sqrt5))/(11-5)-(sqrt11(sqrt11+1))/(sqrt11+1)+sqrt5/10`
`=sqrt11-sqrt5-sqrt11+sqrt5/10`
`=sqrt5/10-sqrt5=(-9sqrt5)/10`
\(\dfrac{6}{\sqrt{11}+\sqrt{5}}-\dfrac{11+\sqrt{11}}{\sqrt{11}+1}+\dfrac{1}{2\sqrt{5}}\)
\(=\sqrt{11}-\sqrt{5}-\sqrt{11}+\dfrac{1}{10}\sqrt{5}\)
\(=-\dfrac{9}{10}\sqrt{5}\)