Thế này à
\(A=\sqrt{3-\sqrt{5}}=\sqrt{\dfrac{6-2\sqrt{5}}{2}}=\sqrt{\dfrac{\left(\sqrt{5}-1\right)^2}{2}}=\dfrac{\sqrt{5}-1}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{2}}{2}\)
Thế này à
\(A=\sqrt{3-\sqrt{5}}=\sqrt{\dfrac{6-2\sqrt{5}}{2}}=\sqrt{\dfrac{\left(\sqrt{5}-1\right)^2}{2}}=\dfrac{\sqrt{5}-1}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{2}}{2}\)
Rút gọn biểu thức sau
\(a.\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\)
\(b.\dfrac{\sqrt{\dfrac{5}{3}}+\sqrt{\dfrac{3}{5}}-2}{\sqrt{\dfrac{5}{3}}-\sqrt{\dfrac{3}{5}}}\)
rút gọn A)\(\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5-3}\right)^2}}\)
B) \(\sqrt{6+2\sqrt{5-\sqrt{\left(2\sqrt{3+1}\right)^2}}}\)
C) \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
Rút gọn \(A=\left(\sqrt{6+\sqrt{20}}-2\sqrt{3-\sqrt{5}}+\sqrt{15-10\sqrt{2}}\right):\left(2+\sqrt{8}\right)\)
rút gọn biểu thức
a).\(\sqrt{2-\sqrt{3}}\)-\(\sqrt{2+\sqrt{3}}\)
b). \(\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
Rút gọn biểu thức sau:
a) A = \(\dfrac{x+\sqrt{5}}{x^2+2x\sqrt{5}+5}\), x ≠ \(-\sqrt{5}\)
b) B = \(\dfrac{a-2\sqrt{a}-3}{a-9}\), a ≥ 0, a ≠ 9
c) C = \(\sqrt{x-1-2\sqrt{x-2}}\)
Bài 1 Rút gọn: D = \(\sqrt{\sqrt{5}-\sqrt{\sqrt{3}-\left(29-12\sqrt{5}\right)}}\)
Rút gọn: \(\left(\sqrt{21}+3\right)\sqrt{5-\sqrt{21}}-\sqrt{24}\)
Rút gọn biểu thức:
a) \(\sqrt{8+4\sqrt{3}}-\sqrt{8-4\sqrt{3}}\)
b) \(\dfrac{\sqrt{21+8\sqrt{5}}}{4+\sqrt{5}}.\sqrt{9-4\sqrt{5}}\)
\(\left(\dfrac{\sqrt{6}-\sqrt{10}}{\sqrt{5}-\sqrt{3}}+3\right)\left(3+\dfrac{2\sqrt{5}+\sqrt{6}}{\sqrt{10}+\sqrt{3}}\right)\)
Rút gọn biểu thức trên
Rút gọn biểu thức sau
\(a.\dfrac{\sqrt{5}-2}{5+2\sqrt{5}}-\dfrac{1}{2+\sqrt{5}}+\dfrac{1}{\sqrt{5}}\)
\(b.\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)
\(c.\dfrac{2\sqrt{3}-4}{\sqrt{3}-1}+\dfrac{2\sqrt{2}-1}{\sqrt{2}-1}-\dfrac{1+\sqrt{6}}{\sqrt{2}+3}\)