Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
Các câu hỏi tương tự
Tính :a ) Sfrac{1}{sqrt{1}sqrt{3}}+frac{1}{sqrt{3}+sqrt{5}}+frac{1}{sqrt{5}+sqrt{7}}+.....+frac{1}{sqrt{2017}+sqrt{2019}}b ) Sfrac{1}{sqrt{2}+sqrt{4}}+frac{1}{sqrt{4}+sqrt{6}}+....+frac{1}{sqrt{100}+sqrt{102}}c ) Sfrac{1}{sqrt{1}+sqrt{2}}+frac{1}{sqrt{2}+sqrt{3}}+.....+frac{1}{sqrt{100}+sqrt{101}}d ) Sfrac{1}{sqrt{3}+sqrt{6}}+frac{1}{sqrt{6}+sqrt{9}}+frac{1}{sqrt{9}+sqrt{12}}+....+frac{1}{sqrt{2016}+sqrt{2019}}
Đọc tiếp
Tính :
a ) \(S=\frac{1}{\sqrt{1}\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+.....+\)\(\frac{1}{\sqrt{2017}+\sqrt{2019}}\)
b ) \(S=\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+....+\frac{1}{\sqrt{100}+\sqrt{102}}\)
c ) \(S=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+.....+\frac{1}{\sqrt{100}+\sqrt{101}}\)
d ) \(S=\frac{1}{\sqrt{3}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{12}}+....+\frac{1}{\sqrt{2016}+\sqrt{2019}}\)
Chứng minh rằng số x0=\(\sqrt{2+\sqrt{2+\sqrt{3}}}-\sqrt{6-3\sqrt{2+\sqrt{3}}}\)là 1 nghiệm của pt x\(x^4+16^2+32=0\)
chứng minh rằng
\(\left(3-2\sqrt{6}-\sqrt{33}\right)\times\left(\sqrt{6}+\sqrt{4}+4\right)=24\)
Rút gọn biểu thứcA frac{sqrt{3}+sqrt{11+6sqrt{2}}-sqrt{5+2sqrt{6}}}{sqrt{2}+sqrt{6+2sqrt{5}}-sqrt{7+2sqrt{10}}}B sqrt{5sqrt{3}+5sqrt{48-10sqrt{7+4sqrt{3}}}}C sqrt{2-sqrt{3}}-sqrt{2+sqrt{3}}D sqrt{28+6sqrt{3}}-sqrt{28-6sqrt{3}}E 6x+sqrt{9x^2-12x+4}F 5x-sqrt{x^2+4x+4}
Đọc tiếp
Rút gọn biểu thức
A = \(\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)
B = \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
C = \(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
D = \(\sqrt{28+6\sqrt{3}}-\sqrt{28-6\sqrt{3}}\)
E = \(6x+\sqrt{9x^2-12x+4}\)
F = \(5x-\sqrt{x^2+4x+4}\)
Chứng minh rằng: \(\sqrt{2\sqrt{3\sqrt{4\sqrt{5...\sqrt{2000}}}}}< 3\)
Chứng minh A,B là số nguyên với:
A = \(\sqrt{6-2\sqrt{5}}\)- \(\sqrt{6+2\sqrt{5}}\)
B= \(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17}-12\sqrt{2}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
Tính \(C=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{7+4\sqrt{3}}-x}{\sqrt[4]{9-4\sqrt{5}}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
tính:\(\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}+\frac{1}{\sqrt{5}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{9}}\)
Rút gọn
1) \(B=\sqrt{\sqrt{7}+6+\sqrt{13-2\sqrt{64-6\sqrt{7}}}}\)
2) \(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
3) \(D=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)