Đặt: \(A=\sqrt{2-\sqrt{3}}\)
\(\Rightarrow\sqrt{2}A=\sqrt{2}.\sqrt{2-\sqrt{3}}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)
\(\Rightarrow A=\dfrac{\sqrt{3}-1}{\sqrt{2}}=\dfrac{\sqrt{6}-\sqrt{2}}{2}\)
\(\sqrt{\dfrac{72.x}{128}}=\dfrac{3}{4}\) Giải phương trình: giúp mik vs
\(\sqrt{3}x^2-\sqrt{1587}x=0\)
\(\sqrt{3-\sqrt{5}}\) : \(\sqrt{2}\)
Giúp mình với
giải giúp mik vs
A=\(\sqrt{4-2\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
B=\(\sqrt{3+\sqrt{8}+\sqrt{3-\sqrt{8}}}\)
Làm tính chia:
a,(\(3\sqrt{x^2y}\)-\(4\sqrt{xy^2}\)+5xy):\(\sqrt{xy}\)
b,(\(\sqrt{a^3b}\)+\(\sqrt{ab}\)-\(3\sqrt{ab^3}\)):\(\sqrt{ab}\)
\(\sqrt{\frac{5+2\sqrt{2}}{5-2\sqrt{6}}}\) + \(\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
\(\sqrt{5}\) - \(\sqrt{3-\sqrt{29-6\sqrt{20}}}\)
GIÚP MK VS MAI MK NỘP RÙI
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\(D=\sqrt{2+\sqrt{3}}+\sqrt{14-5\sqrt{3}}+\sqrt{2}\)
B3: làm phép chia :
\(\sqrt{\dfrac{a-1}{a+2}}\div\sqrt{\dfrac{a+2}{a^3-3a^2+3a-1}}\) với a>1
bài 1 Rút gọn biểu thức:A=\(\sqrt{1+\dfrac{1}{a^2}+\dfrac{1}{\left(1+a\right)^2}}\)với a>0
2) Tính giá trị của tổng:
a) B =\(\sqrt{1+\dfrac{1}{1^2}+\dfrac{1}{2^2}}\)+ \(\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}\)+\(\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}\)+....+\(\sqrt{1+\dfrac{1}{2011^2}+\dfrac{1}{2012^2}}\)
\(\dfrac{\sqrt{8-4\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{4\cdot2-4\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{4}\cdot\sqrt{2-\sqrt{3}}}{\sqrt{2}}=\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\)
