b) Ta có: \(\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}+\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)
\(=\sqrt{\dfrac{\left(3+\sqrt{5}\right)^2}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}}+\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}}\)
\(=\dfrac{3+\sqrt{5}}{2}+\dfrac{3-\sqrt{5}}{2}\)
\(=\dfrac{3+3}{2}=\dfrac{6}{2}=3\)
không dùng máy tính , tính giá trị của các biểu thức sau
1)\(\left(1+\sqrt{2}+\sqrt{3}\right)\cdot\left(1+\sqrt{2}+\sqrt{3}\right)\)
2)\(\dfrac{1}{\sqrt{2}+1}-\dfrac{\sqrt{8}-\sqrt{10}}{2-\sqrt{5}}\)
3)\(\dfrac{2+\sqrt{3}}{\sqrt{7-4\sqrt{3}}}-\dfrac{2-\sqrt{3}}{\sqrt{7+4\sqrt{3}}}\)
4)\(\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)-\dfrac{\sqrt{7-4\sqrt{3}}}{\sqrt{3}-2}\)
5)\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
6)\(\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}\)
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) \(\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
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
\(\dfrac{1}{3-\sqrt{8}}-\dfrac{1}{\sqrt{8}-\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-2}\)
1) rút gọn biểu thức sau :
a) \(\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\) b) \(\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\) c ) \(\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
d) \(\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\) e) \(\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{y}}\) ( với x>0 , y>0 )
f) \(\sqrt{8-2\sqrt{15}}+\sqrt{5}+\sqrt{3}\) g) \(\sqrt{9-2\sqrt{4}}-\sqrt{9+2\sqrt{14}}\)
cho M= \(\sqrt{4+\sqrt{7}}-\sqrt{\sqrt{7}+\sqrt{3}}\) chứng minh M=\(\sqrt{2}\)
cho M=\(\dfrac{\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}}{\sqrt{\sqrt{7}-2}}\) chứng minh M=-\(\sqrt{2}\)
CHO M=\(\sqrt{\dfrac{3\sqrt{3}-4}{2\sqrt{3}+1}}\)+\(\sqrt{\dfrac{\sqrt{3}+4}{5-2\sqrt{3}}}\) chứng minh M=\(\sqrt{6}\)
giúp mk vs mk cần gấp lắm
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) \(\dfrac{1}{\sqrt{5}-\sqrt{3}}\)\(-\)\(\dfrac{1}{\sqrt{5+\sqrt{3}}}\)
1) Rút gọn :
a) \(A=\sqrt{\left(\sqrt{3}-\sqrt{5}\right)}^2-\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}\)
b) B = \(\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}\)
c) C = \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
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=\(\dfrac{1+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{1-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
