Đưa biểu thức trong căn về dạng hình phương của một tổng hoặc một hiệu:
a/ \(\sqrt{7-2\sqrt{10}}-\sqrt{6-2\sqrt{5}}\)
b/ \(\sqrt{33-12\sqrt{6}}+\sqrt{15-6\sqrt{6}}\)
c/\(\sqrt{9-4\sqrt{5}}+\sqrt{12-2\sqrt{35}}\)
d/ \(\sqrt{4-2\sqrt{3}}+\sqrt{28-10\sqrt{3}}\)
e/ \(\frac{\sqrt{5}-\sqrt{15}}{1-\sqrt{3}}-\sqrt{21+4\sqrt{5}}\)
a. P= (\(3+\sqrt{2}+\sqrt{6}\))(\(\sqrt{6-3\sqrt{3}}\))
b. A=(\(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\)): (\(\sqrt{6}+11\))
c. B= \(\frac{\sqrt{8-2\sqrt{12}}}{\sqrt{3}-1}\)-\(\sqrt{8}\)
d. C= \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
đ. D=\(\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
e. E= \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
ê. G= \(\sqrt{4+5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
g. H=\(\frac{2\sqrt{4+\sqrt{5+21+\sqrt{80}}}}{\sqrt{10}-\sqrt{2}}\)
i. I=\(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}+\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}\)
k. K=\(\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
giúp mk tính
a,\(\sqrt{5}-\sqrt{48}+5\sqrt{27}-\sqrt{45}\)
b,(\(\sqrt{5}+\sqrt{2}\)) (\(3\sqrt{2}-1\))
c,\(3\sqrt{50}-2\sqrt{75}-4\dfrac{\sqrt{54}}{\sqrt{3}}-3\sqrt{\dfrac{1}{3}}\)
d, \(\sqrt{\left(\sqrt{3}-3\right)^2}+\sqrt{4-2\sqrt{3}}\)
e, \(\sqrt{48-2\sqrt{135}}-\sqrt{45}+\sqrt{18}\)
f, \(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\dfrac{6}{2-\sqrt{10}}-\dfrac{20}{\sqrt{10}}\)
bài 2
a, \(\sqrt{9-4\sqrt{5}}\)
b,\(2\sqrt{3}+\sqrt{48}-\sqrt{75}-\sqrt{243}\)
c\(\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
d, \(\sqrt{3+2\sqrt{2}}-\sqrt{6-4\sqrt{2}}\)
e,\(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)+\(\dfrac{\sqrt{3}+\sqrt{5}}{\sqrt{5}-\sqrt{3}}-\dfrac{\sqrt{5}+1}{\sqrt{5}-1}\)
f, \(\sqrt{5\sqrt{3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
Trục căn ở mẫu:
\(a)\frac{5}{\sqrt{10}}\\ b)\frac{-2}{1-\sqrt{5}}\\ c)\frac{4}{\sqrt{3}+\sqrt{2}}\\ d)\frac{1}{3-2\sqrt{2}}\\ e)\frac{6-\sqrt{6}}{1-\sqrt{6}}\\ g)\frac{3\sqrt{2}-2\sqrt{3}}{2\left(\sqrt{3}-\sqrt{2}\right)}\\ h)\frac{\sqrt{3}-3}{\sqrt{3}-1}\\ i)\frac{\sqrt{15}}{5\sqrt{3}+3\sqrt{5}}\)
Bài 1: Rút gọn biểu thức
a) \(A=\sqrt{26+15\sqrt{3}}\)
b) \(B=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
c) \(C=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
d) \(D=\left(\sqrt{6}-2\right)\left(5+\sqrt{24}\right)\sqrt{5-\sqrt{24}}\)
e) \(E=\left(\sqrt{10}-\sqrt{2}\right)\left(\sqrt{3+\sqrt{5}}\right)\)
f) \(F=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
g) \(G=\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)
h) \(H=\frac{\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\)
Rút gọn
H=\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
F=\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
G=\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
E=\(\frac{2\sqrt{3+\sqrt{5-13+\sqrt{48}}}}{\sqrt{6}+\sqrt{2}}\)
D=\(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
Z=\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10-2\sqrt{5}}}\)
Trục căn thức ở mẫu
a)\(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}-\sqrt{7}}\)
b)\(\dfrac{5\sqrt{5}+3\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
c)\(\dfrac{3+4\sqrt{3}}{\left(\sqrt{6}+\sqrt{2}\right)-\sqrt{5}}\)
Thực hiện phép tính:
a) \(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
b)\(\sqrt{24-8\sqrt{5}}+\sqrt{9+4\sqrt{5}}\)
c)\(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)
d)\(\sqrt{41+12\sqrt{5}}-\sqrt{46-6\sqrt{5}}\)
e)\(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
f)\(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)
g)\(\sqrt{43+24\sqrt{3}}-\sqrt{49-\sqrt{8\sqrt{3}}}\)
h)\(\sqrt{53-20\sqrt{7}}-\sqrt{64+6\sqrt{7}}\)
So sánh 2 số: \(R=\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(S=\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)