rút gọn biểu thức
a, \(\dfrac{1}{\sqrt{7-\sqrt{24}+1}}-\dfrac{1}{\sqrt{7+\sqrt{24}+1}}\)
b,\(\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}+\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)
c,\(\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4}+\sqrt{7}}+\dfrac{4-\sqrt{7}}{3\sqrt{7}-\sqrt{4}-\sqrt{7}}\)
Bài: Rút gọn biểu thức:
a, \(\dfrac{1}{\sqrt{7-\sqrt{24}+1}}-\dfrac{1}{\sqrt{7-\sqrt{24}-1}}\)
b, \(\sqrt{\dfrac{4}{9-4\sqrt{5}}}-\sqrt{\dfrac{4}{9+4\sqrt{5}}}\)
rút gọn biểu thức
K=\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{25\sqrt{24}+24\sqrt{25}}\)
M=\(\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
N=\(\dfrac{1+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{1-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
a)cho a>b>0 chứng minh rằng : \(\dfrac{1}{a+b}\le\dfrac{1}{2\sqrt{ab}}\)
b) Chứng minh \(\dfrac{\sqrt{2}-\sqrt{1}}{3}+\dfrac{\sqrt{3}-\sqrt{2}}{5}+\dfrac{\sqrt{4}-\sqrt{3}}{7}+...+\dfrac{\sqrt{2011}-\sqrt{2010}}{4021}< \dfrac{1}{2}\)
giúp mk vs
1)\(\sqrt{12}\)\(-\)\(\sqrt{27}\)\(+\)\(\sqrt{48}\)
2)(\(\sqrt{24}+\sqrt{20}-\sqrt{80}\))\(\div\)5
3)2\(\sqrt{27}-\sqrt{\dfrac{16}{3}}\)\(-\)\(\sqrt{48}-\)\(\sqrt{8\dfrac{1}{3}}\)
4) \(\dfrac{1}{\sqrt{5}-\sqrt{3}}\)\(-\)\(\dfrac{1}{\sqrt{5+\sqrt{3}}}\)
Tính:
a) \(\dfrac{\sqrt{7}-5}{2}-\dfrac{6-2\sqrt{7}}{4}+\dfrac{6}{\sqrt{7}-2}-\dfrac{5}{4+\sqrt{7}}\)
b) \(\dfrac{2}{\sqrt{6}-2}+\dfrac{2}{\sqrt{6}+2}+\dfrac{5}{\sqrt{6}}\)
c) \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{1}{\sqrt{6}}}\)
d) \(\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
chứng minh bất đẳng thức:
\(\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+\dfrac{1}{\sqrt{9}+\sqrt{11}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}\)<\(\dfrac{9}{4}\)
Thực hiện phép tính:
a) \(\sqrt{24+8\sqrt{5}}-\sqrt{9-4\sqrt{5}}.\)
b) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\).
c) \(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+....+\dfrac{1}{\sqrt{99}+\sqrt{100}}.\)
Bài 1. Tính
a) A= \(\left[\dfrac{6+\sqrt{20}}{3+\sqrt{5}}+\dfrac{\sqrt{14}-\sqrt{2}}{\sqrt{7}-1}\right]\) : (2+ \(\sqrt{2}\))
b) B= \(\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-\dfrac{11}{2\sqrt{3}+1}\)
Bài 2.
Cho A= \(\left(\dfrac{\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}-\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}\right).\dfrac{\sqrt{3}-1}{3\sqrt{2}-\sqrt{6}}\)
Chứng minh A là số nguyên.
