\(\sqrt{11-6\sqrt{2}}+\sqrt{11+6\sqrt{2}}\)
= \(\sqrt{9-2.3.\sqrt{2}+2}+\sqrt{9+6\sqrt{2}+2}\)
= \(\sqrt{\left(3-\sqrt{2}\right)^2}+\sqrt{\left(3+\sqrt{2}\right)^2}\)
= 3 - \(\sqrt{2}\) + 3 + \(\sqrt{2}\) = 6
Chứng minh rằng:
a)\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{5-2\sqrt{6}}\right)}{9\sqrt{3}-11\sqrt{2}}\) là số nguyên
b)\(\left(\sqrt{3}-1\right).\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
Rút gọn căn bậc hai theo hằng đẳng thức:
a)\(\left(4\sqrt{2}+\sqrt{30}\right).\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)
b)\(2.\left(\sqrt{10}-\sqrt{2}\right).\left(4+\sqrt{6-2\sqrt{5}}\right)\)
c)\(\left(7+\sqrt{14}\right).\sqrt{9-2\sqrt{14}}\)
d)\(\sqrt{\dfrac{289+4\sqrt{72}}{16}}\)
e) \(\left(\sqrt{21}+7\right).\sqrt{10-2\sqrt{21}}\)
f)\(\sqrt{2-\sqrt{3}.\left(\sqrt{6}+\sqrt{2}\right)}\)
g) \(\sqrt{2}\sqrt{8+3\sqrt{7}}\)
h) \(\sqrt{11+6\sqrt{2}}\)
Tính:
a)\(\dfrac{-3}{5}\) . \(\sqrt{\left(-0,5\right)^{ }2}\)
b)4.\(\sqrt{\left(-3\right)^6}\) + 5\(\sqrt{\left(-2\right)^4}\)
c)\(\sqrt{\left(1-\sqrt{7}\right)^2}\) +\(\sqrt{7}\)
d)\(\sqrt{11+6\sqrt{2}}\) - \(\sqrt{11-6\sqrt{2}}\)
-GIÚP MÌNH VỚI Ạ-
\(\sqrt{\left(\sqrt{2}-3\right)^2}.\sqrt{11+6\sqrt{2}}\)
1.Tính :
a.\(\sqrt{11+6\sqrt{2}}+\sqrt{19+6\sqrt{2}}\)
b.\(-\sqrt{2\left(2-\sqrt{3}\right)}+\sqrt{2\left(2+\sqrt{3}\right)}\)
c.\(\frac{\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}}{2\sqrt{3}}\)
1.Tính:
a.\(\sqrt{11+6\sqrt{2}-\sqrt{19+6\sqrt{2}}}\)
b.\(-\sqrt{2\left(2-\sqrt{3}\right)}+\sqrt{2\left(2+\sqrt{3}\right)}\)
c.\(\frac{\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}}{2\sqrt{3}}\)
\(\sqrt{\left(11-6\sqrt{2}\right)^2}\)Tách như thế nào vậy mn ??
\(\sqrt{7-\sqrt{24}}-\dfrac{\sqrt{50}-5}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11-\sqrt{120}\right)\left(11+2\sqrt{30}\right)^2}\)
Rút gọn giùm mình với ạ
Rút gọn
\(C=\left(\sqrt{12+2\sqrt{14+2\sqrt{13}}}-\sqrt{12+2\sqrt{11}}\right)\left(\sqrt{11}+\sqrt{13}\right)\)
