Ta có: \(\sqrt{27}-6\sqrt{\dfrac{1}{3}}-\sqrt{5-2\sqrt{6}}\)
= \(3\sqrt{3}-2\sqrt{3}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
= \(\sqrt{3}-\sqrt{3}+\sqrt{2}\)
= \(\sqrt{2}\)
Ta có: \(\sqrt{27}-6\sqrt{\dfrac{1}{3}}-\sqrt{5-2\sqrt{6}}\)
= \(3\sqrt{3}-2\sqrt{3}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
= \(\sqrt{3}-\sqrt{3}+\sqrt{2}\)
= \(\sqrt{2}\)
CM: \(\left(\dfrac{2}{\sqrt{6}-1}+\dfrac{3}{\sqrt{6}-2}+\dfrac{3}{\sqrt{6}-3}\right).\dfrac{5}{9\sqrt{6}+4}=\dfrac{1}{2}\)
Chứng minh các đẳng thức:
a) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)=1
b)\(\dfrac{\left(5+2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{\sqrt{2}+\sqrt{3}}\)-1 =0
c) \(\sqrt{26+15\sqrt{3}}+\sqrt{26-15\sqrt{3}}-5\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{6}}{2}\)
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, \(\sqrt{5\sqrt{3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
thực hiện phép tính
a)\(\sqrt{80}-\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{3\dfrac{1}{5}}\)
b)\(\dfrac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}+\dfrac{3+6\sqrt{3}}{\sqrt{3}}-\dfrac{13}{\sqrt{3}+4}\)
thực hiện phép tính sau: \(\dfrac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(\dfrac{\sqrt{3}}{2-\sqrt{6}}+\dfrac{\sqrt{3}}{2+\sqrt{6}}\right)-\dfrac{1}{\sqrt{2}}\)
Tính:
\(a.\) \(A=\sqrt{12}-2\sqrt{48}+\dfrac{7}{5}\sqrt{75}\)
\(b.\) \(B=\sqrt{14-6\sqrt{5}}+\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(c.\) \(C=\left(\sqrt{6}-\sqrt{2}\right)\sqrt{2+\sqrt{3}}\)
\(d.\) \(D=\dfrac{5+\sqrt{5}}{\sqrt{5}+2}+\dfrac{\sqrt{5}-5}{\sqrt{5}}-\dfrac{11}{2\sqrt{5}+3}\)
chứng minh \(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\dfrac{1}{\sqrt{5}+\sqrt{6}}+...+\dfrac{1}{\sqrt{2023}+\sqrt{2024}}>22\)
\(\dfrac{6-2\sqrt{3}}{\sqrt{3}-1}-\dfrac{2}{\sqrt{5}+\sqrt{3}}-\sqrt{17+4\sqrt{15}}\)
Tính :\(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\dfrac{1}{\sqrt{5}+\sqrt{6}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)