Ta có: \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\)
\(=\dfrac{\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}-2}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+1-\sqrt{5}+1-2}{\sqrt{2}}\)
=0
Rút gọn : \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
Rút gọn biểu thức : \(\sqrt{4+\sqrt{10+2\sqrt{5}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}}\)
bài 1 rút gọn
a \(A=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
b\(B=\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}\)
c\(C=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\) d\(D=\sqrt{2+\sqrt{3}}+\sqrt{14-5\sqrt{3}}+\sqrt{2}\)
Rút gọn
\(\dfrac{2+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
Cảm ơn
Rút gọn biểu thức
\(\dfrac{\left(\sqrt{3}-\sqrt{5}\right)^2+4\sqrt{15}}{\sqrt{3}+\sqrt{5}}\)
Rút gọn các biểu thức
A= \(\sqrt{\left(\sqrt{2}-1\right)^2}\)-\(\sqrt{3+2\sqrt{2}}\)
B= \(\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}\)
C= \(\sqrt{\left(2\sqrt{5}-7\right)^2}\)-\(\sqrt{45-20\sqrt{5}}\)
D= \(\sqrt{\left(3-\sqrt{5}\right)^2}\)+\(\sqrt{5}\)
BT1: Tính
a, \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\)
b, \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{7}\)
BT2: Rút gọn
\(3x-\sqrt{27}+\frac{\sqrt{x^3+3x^2}}{\sqrt{x+3}}\) ( x ≥ 0 )
BT1: Tính
a, \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{12}\)
b, \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{7}\)
BT2: Rút gọn
\(3x-\sqrt{27}+\frac{\sqrt{x^3+3x^2}}{\sqrt{x+3}}\) ( x ≥ 0 )
Rút gọn các căn thức sau
\(\sqrt{12-3\sqrt{7}}\)
\(\sqrt{2-\sqrt{3}}\)
\(\sqrt{3-\sqrt{5}}\)
\(\sqrt{7-3\sqrt{5}}\)
