I : Rút gọn
D=\(\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}-\sqrt{\frac{3+\sqrt{5}}{3-\sqrt{5}}}\)
\(E=\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
help me !!!
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I : Rút gọn
\(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2019\sqrt{2020}+2020\sqrt{2019}}\)
help me !!!
Lời giải:
Xét số hạng tổng quát:
\(\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n(n+1)}(\sqrt{n}+\sqrt{n+1})}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Do đó:
\(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2019}}-\frac{1}{\sqrt{2020}}\)
\(=1-\frac{1}{\sqrt{2020}}\)
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Trục căn ở mẫu:
\(a)\frac{5}{\sqrt{10}}\\ b)\frac{-2}{1-\sqrt{5}}\\ c)\frac{4}{\sqrt{3}+\sqrt{2}}\\ d)\frac{1}{3-2\sqrt{2}}\\ e)\frac{6-\sqrt{6}}{1-\sqrt{6}}\\ g)\frac{3\sqrt{2}-2\sqrt{3}}{2\left(\sqrt{3}-\sqrt{2}\right)}\\ h)\frac{\sqrt{3}-3}{\sqrt{3}-1}\\ i)\frac{\sqrt{15}}{5\sqrt{3}+3\sqrt{5}}\)
Rút gọn
a)\(\frac{2\sqrt{5\:\:\:}-4\sqrt{10}}{3\sqrt{10}}\)
b)\(\frac{6\sqrt{6}-2\sqrt{12}+3-\sqrt{2}}{2\sqrt{6}+1}\)
c)\(\frac{5\sqrt{7}-4\sqrt{35}+7\sqrt{5}}{\sqrt{35}}\)
d)\(\left(\sqrt{3}-1\right)\sqrt{2\sqrt{19+8\sqrt{3}}-4}\)
Rút gọn biểu thức sau
\(A=\frac{\sqrt{x+2}}{\sqrt{x-3}}-\frac{\sqrt{x+1}}{\sqrt{x-2}}-\frac{3\sqrt{x-3}}{x-5\sqrt{x+6}}\)
Rút gọn
\(\sqrt{45a}-2\sqrt{\frac{4a}{3}}+\frac{\sqrt{18a}}{\sqrt{6}}+\sqrt{5\frac{1}{3}a}\)
tính:a)frac{1}{1+sqrt{5}}+frac{1}{1-sqrt{5}}b)frac{sqrt{15}-sqrt{12}}{sqrt{5}-2}+frac{1}{2-sqrt{3}}c)frac{2}{sqrt{5}+1}+sqrt{frac{2}{3-sqrt{5}}}-5sqrt{frac{1}{5}}d)left(frac{5}{sqrt{15}-sqrt{10}}-frac{3sqrt{5}-5sqrt{3}}{sqrt{3}-sqrt{5}}right)^2e)frac{2}{sqrt{3}-sqrt{5}}+frac{3-2sqrt{3}}{sqrt{3}-2}
Đọc tiếp
tính:
a)\(\frac{1}{1+\sqrt{5}}+\frac{1}{1-\sqrt{5}}\)
b)\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}\)
c)\(\frac{2}{\sqrt{5}+1}+\sqrt{\frac{2}{3-\sqrt{5}}}-5\sqrt{\frac{1}{5}}\)
d)\(\left(\frac{5}{\sqrt{15}-\sqrt{10}}-\frac{3\sqrt{5}-5\sqrt{3}}{\sqrt{3}-\sqrt{5}}\right)^2\)
e)\(\frac{2}{\sqrt{3}-\sqrt{5}}+\frac{3-2\sqrt{3}}{\sqrt{3}-2}\)
a,frac{2sqrt{2}}{sqrt{2sqrt{2}+1}+1}-frac{2sqrt{2}}{sqrt{2sqrt{2}+1}-1}
b,frac{sqrt{15}-sqrt{12}}{sqrt{5}-2}+frac{1}{2-sqrt{3}}
c,frac{2}{sqrt{3}-sqrt{5}}+frac{3-2sqrt{3}}{sqrt{3}-2}
d,frac{-4}{sqrt{7}-sqrt{5}}+frac{1}{sqrt{3}-1}+frac{4-2sqrt{5}}{sqrt{5}-2}
e,frac{6}{sqrt{5}-1}+frac{7}{1-sqrt{3}}-frac{2}{sqrt{3}-sqrt{5}}
Đọc tiếp
a,\(\frac{2\sqrt{2}}{\sqrt{2\sqrt{2}+1}+1}-\frac{2\sqrt{2}}{\sqrt{2\sqrt{2}+1}-1}\)
b,\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}\)
c,\(\frac{2}{\sqrt{3}-\sqrt{5}}+\frac{3-2\sqrt{3}}{\sqrt{3}-2}\)
d,\(\frac{-4}{\sqrt{7}-\sqrt{5}}+\frac{1}{\sqrt{3}-1}+\frac{4-2\sqrt{5}}{\sqrt{5}-2}\)
e,\(\frac{6}{\sqrt{5}-1}+\frac{7}{1-\sqrt{3}}-\frac{2}{\sqrt{3}-\sqrt{5}}\)
Bài 1: Tính
1, Aleft(1-frac{5+sqrt{5}}{1+sqrt{5}}right).left(frac{5-sqrt{5}}{1-sqrt{5}}-1right)
2, Bleft(frac{3sqrt{125}}{15}-frac{10-4sqrt{6}}{sqrt{5}-2}right).frac{1}{sqrt{5}}
3, Cleft(frac{sqrt{1000}}{100}-frac{5sqrt{2}-2sqrt{5}}{2sqrt{5}-8}right).frac{sqrt{10}}{10}
4, Dfrac{1}{sqrt{49+20sqrt{6}}}-frac{1}{sqrt{49-20sqrt{6}}}+frac{1}{sqrt{7-4sqrt{3}}}
5, Efrac{1}{sqrt{4-2sqrt{3}}}-frac{1}{sqrt{7-sqrt{48}}}+frac{3}{sqrt{14-6sqrt{5}}}
6, Ffrac{1}{sqrt{2}-sqrt{3}}sqrt{frac{3sqrt{2}-2sqrt{3}...
Đọc tiếp
Bài 1: Tính
1, \(A=\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right).\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
2, \(B=\left(\frac{3\sqrt{125}}{15}-\frac{10-4\sqrt{6}}{\sqrt{5}-2}\right).\frac{1}{\sqrt{5}}\)
3, \(C=\left(\frac{\sqrt{1000}}{100}-\frac{5\sqrt{2}-2\sqrt{5}}{2\sqrt{5}-8}\right).\frac{\sqrt{10}}{10}\)
4, \(D=\frac{1}{\sqrt{49+20\sqrt{6}}}-\frac{1}{\sqrt{49-20\sqrt{6}}}+\frac{1}{\sqrt{7-4\sqrt{3}}}\)
5, \(E=\frac{1}{\sqrt{4-2\sqrt{3}}}-\frac{1}{\sqrt{7-\sqrt{48}}}+\frac{3}{\sqrt{14-6\sqrt{5}}}\)
6, \(F=\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
7, \(G=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}-\sqrt{11-2\sqrt{10}}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}+\sqrt{12+8\sqrt{2}}}}\)
Afrac{2sqrt{8}-sqrt{12}}{sqrt{18}-sqrt{48}}-frac{sqrt{5}+sqrt{27}}{sqrt{30}+sqrt{162}}Bsqrt{frac{2-sqrt{3}}{2+sqrt{3}}}+sqrt{frac{2+sqrt{3}}{2-sqrt{3}}}Cfrac{sqrt{3-sqrt{5}}left(3+sqrt{5}right)}{sqrt{10}+sqrt{2}}Dfrac{1}{sqrt{2}+sqrt{2+sqrt{3}}}+frac{1}{sqrt{2}-sqrt{2-sqrt{3}}}Efrac{left(sqrt{5+2}right)^2-8sqrt{5}}{2sqrt{5}-4}
Đọc tiếp
\(A=\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
\(B=\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(C=\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
\(D=\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(E=\frac{\left(\sqrt{5+2}\right)^2-8\sqrt{5}}{2\sqrt{5}-4}\)
Phần d mình sửa lại đề nha : \(\frac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{4}-4}\)
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bn xem lại đề câu d đi sao mẫu lại bằng 0 rồi
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