Nhân B với \(\sqrt{2}\) ta được:
\(B.\sqrt{2}\)=\(\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right).\sqrt{2}\)
= \(\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\)
= \(\sqrt{7+2\sqrt{7}+1}-\sqrt{7-2\sqrt{7}+1}\)
= \(\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}\)
= \(\sqrt{7}+1-\sqrt{7}+1\)
= 2
Tính: a. \(\left(\sqrt{10}+\sqrt{2}\right)\cdot\left(6-2\sqrt{5}\right)\cdot\sqrt{3+\sqrt{5}}\)
b. \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\)
c. \(\sqrt{3,5-\sqrt{6}}+\sqrt{3,5+\sqrt{6}}\)
d, \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{7}\)
e, \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
Tính giá trị các biểu thức:
a/ \(\sqrt{7+\sqrt{7^2-4}}-\sqrt{7-\sqrt{7^2-4}}\)
b/ \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
c/ \(\left(\dfrac{1}{2+\sqrt{3}}-\dfrac{12}{3+\sqrt{3}}+\dfrac{26}{4-\sqrt{3}}\right)\cdot\left(4-3\sqrt{3}\right)\)
Không dùng máy tính hãy so sánh:
\(A=\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(B=\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)
\(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)
\(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)
\(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)
\(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2^6\right)}\)
rút gọn:giải chi tiết hộ mình nha
Rút gọn biểu thức:
a) \(\sqrt{9-2\sqrt{14}}\)
b) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
c) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
Tính
\(a.\sqrt{5-2\sqrt{5}}+\sqrt{5+2\sqrt{5}}\)
\(b.\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
\(c.\sqrt{13-4\sqrt{3}}+\sqrt{13+4\sqrt{3}}\)
giai giups nhanh nha,RUT GON
A=\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{7}\)
B=\(\sqrt{6.5+\sqrt{12}}+\sqrt{6.5-\sqrt{12}}+2\sqrt{6}\)
C=\(\sqrt{46+\sqrt{6\sqrt{5}}}-\sqrt{29-12\sqrt{5}}\)
D=\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
* Rút gọn:
a.\(\sqrt{\left(\sqrt{7}-4\right)^2+\sqrt{7}}\)
b.\(\sqrt{81a}-\sqrt{144a}+\sqrt{36a}\) với a≥0
Tính: a, \(\sqrt{2006+2\sqrt{2005}}-\sqrt{2006-2\sqrt{2005}}\)
b, \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
