=\(\sqrt{3\sqrt{2}-12\sqrt{2}}\)
=\(\sqrt{\sqrt{2}.\left(3-12\right)}\)
=\(\sqrt{\sqrt{2}.-9}\)
vậy pt vô nghiệm vì trong dấu căn không có âm(\(\sqrt{\sqrt{2}-9}\le0\)
Thu gọn các biểu thức:
\(B=\sqrt{18-4\sqrt{15}-4\sqrt{3}+2\sqrt{5}}-\sqrt{13-4\sqrt{3}}\)
\(C=\sqrt{28+12\sqrt{2}-4\sqrt{6}-12\sqrt{3}}-\sqrt{18+\sqrt{128}-\sqrt{96}-2\sqrt{48}}\)
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(\sqrt{11+2\sqrt{18}}\)
\(\sqrt{7+2\sqrt{10}}\)
\(\sqrt{7+4\sqrt{3}}\)
\(\sqrt{12-2\sqrt{55}}\)
rút gọn biểu thức
A=2015+\(\sqrt{36}\)-\(\sqrt{25}\)
B=5\(\sqrt{8}\)+\(\sqrt{50}\)-2\(\sqrt{18}\)
C=\(\sqrt{27}\)-2\(\sqrt{12}\)-\(\sqrt{75}\)
D=\(\sqrt{12}\)+\(\sqrt{27}\)-\(\sqrt{48}\)
\(A=2\sqrt{27}-3\sqrt{12}+\sqrt{98}-\sqrt{18}\)
4) A = 3.\(\sqrt{2}\)+ 4.\(\sqrt{8}\)-\(\sqrt{18}\)
5) ( \(\sqrt{12}-\sqrt{48}-\sqrt{108}-\sqrt{192}\) ) :2\(\sqrt{3}\)
\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
\(\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18}-\sqrt{128}}}}\)
a,\(\sqrt{8+2\sqrt{15}}\) -\(\sqrt{6+2\sqrt{15}}\)
b, \(\sqrt{17-2\sqrt{72}}-\sqrt{19+2\sqrt{18}}\)
c, \(\sqrt{8-2\sqrt{7}}+\sqrt{8+2\sqrt{7}}\)
d, \(\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}\)
e, \(\sqrt{10-2\sqrt{21}}-\sqrt{9-2\sqrt{14}}\)
rút gọn
\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
