\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+.......+\frac{1}{1001\sqrt{1000}}+\frac{1}{1002\sqrt{1001}}< \sqrt{2}\)
Những câu hỏi liên quan
Tính nhanh : \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt[1]{2}+\sqrt[2]{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt[3]{4}+\sqrt[4]{5}}+...+\frac{1}{\sqrt{999}+\sqrt{1000}}+\frac{1}{\sqrt[999]{1000}+\sqrt[1000]{1001}}\)
TÍnh giá trị biểu thức sau:
\(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+........+\frac{1}{1000\sqrt{999}+999\sqrt{1000}}\)
\(A=\frac{1}{\sqrt{2.1}\left(\sqrt{2}+\sqrt{1}\right)}+\frac{1}{\sqrt{2.3}\left(\sqrt{3}+\sqrt{2}\right)}+\frac{1}{\sqrt{3.4}\left(\sqrt{4}+\sqrt{3}\right)}+...+\frac{1}{\sqrt{999.1000}\left(\sqrt{1000}+\sqrt{999}\right)}\)
\(A=\frac{\sqrt{2}-\sqrt{1}}{\sqrt{2.1}\left(2-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{2.3}\left(3-2\right)}+\frac{\sqrt{4}-\sqrt{3}}{\sqrt{3.4}\left(4-3\right)}+...+\frac{\sqrt{1000}-\sqrt{999}}{\sqrt{999.1000}\left(1000-999\right)}\)
\(A=\frac{\sqrt{2}}{\sqrt{2.1}}-\frac{\sqrt{1}}{\sqrt{2.1}}+\frac{\sqrt{3}}{\sqrt{2.3}}-\frac{\sqrt{2}}{\sqrt{2.3}}+\frac{\sqrt{4}}{\sqrt{3.4}}-\frac{\sqrt{3}}{\sqrt{3.4}}+...+\frac{\sqrt{1000}}{\sqrt{999.1000}}-\frac{\sqrt{999}}{\sqrt{1000.999}}\)
\(A=\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{999}}-\frac{1}{\sqrt{1000}}\)
\(A=\frac{1}{1}-\frac{1}{\sqrt{1000}}=\frac{\sqrt{1000}-1}{\sqrt{1000}}=\frac{10\sqrt{10}-1}{10\sqrt{10}}\)
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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}...
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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, \(G=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}-\sqrt{11-2\sqrt{10}}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}+\sqrt{12+8\sqrt{2}}}}\)
Tính giá trị biểu thức:text{a) }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{2010}+sqrt{2011}}text{b) }frac{1}{2sqrt{1}+1sqrt{2}}+frac{1}{3sqrt{2}+2sqrt{3}}+frac{1}{4sqrt{3}+3sqrt{4}}+...+frac{1}{121sqrt{120}+120sqrt{121}}text{c) }sqrt{1+frac{1}{1^2}+frac{1}{2^2}}+sqrt{1+frac{1}{2^2}+frac{1}{3^2}}+sqrt{1+frac{1}{3^2}+frac{1}{4^2}}+...sqrt{+1+frac{1}{2010^2}+frac{1}{2011^2}}
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Tính giá trị biểu thức:
\(\text{a) }\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+...+\frac{1}{\sqrt{2010}+\sqrt{2011}}\)
\(\text{b) }\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{121\sqrt{120}+120\sqrt{121}}\)
\(\text{c) }\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...\sqrt{+1+\frac{1}{2010^2}+\frac{1}{2011^2}}\)
Rút gọn biểu thức 1) frac{sqrt{5+2sqrt{6}}+sqrt{8+2sqrt{15}}}{sqrt{7+2sqrt{10}}}2) left(2+frac{3+sqrt{3}}{sqrt{3}+1}right)left(2+frac{3-sqrt{3}}{sqrt{3}-1}right):left(sqrt{5}-2right)3) left(frac{15}{sqrt{6}+1}+frac{4}{sqrt{6}-2}-frac{12}{3-sqrt{6}}right).left(sqrt{6}+11right)4) frac{1}{1+sqrt{2}}+frac{1}{sqrt{2}+sqrt{3}}+frac{1}{sqrt{3}+sqrt{4}}+...+frac{1}{sqrt{99}+sqrt{100}}5) frac{1}{1-sqrt{2}}-frac{1}{sqrt{2}-sqrt{3}}+frac{1}{sqrt{3}-sqrt{4}}-...-frac{1}{sqrt{98}-sqrt{99}}+frac{1}{sqrt{99}-s...
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Rút gọn biểu thức
1) \(\frac{\sqrt{5+2\sqrt{6}}+\sqrt{8+2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}\)
2) \(\left(2+\frac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\frac{3-\sqrt{3}}{\sqrt{3}-1}\right):\left(\sqrt{5}-2\right)\)
3) \(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right).\left(\sqrt{6}+11\right)\)
4) \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
5) \(\frac{1}{1-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-...-\frac{1}{\sqrt{98}-\sqrt{99}}+\frac{1}{\sqrt{99}-\sqrt{100}}\)
6) \(\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
7)\(\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{3}{2}}+2\right)\left(\frac{\sqrt{2}+\sqrt{3}}{4\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}+\sqrt{3}}\right)\left(24+8\sqrt{6}\right)\left(\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}}+\frac{\sqrt{3}}{\sqrt{2}-\sqrt{3}}\right)\)
Câu 1,2,3 Ez quá rồi :3
Câu 4:
Tổng quát:
\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v
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Câu 5 ko khác câu 4 lắm :v
Câu 5:
Tổng quát:
\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v
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Sao làm hổng ai bảo đú.n/g vậy :(((
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Tính
a)\(\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+....+\frac{1}{\sqrt{1936}+\sqrt{1935}}\)
b)\(\frac{1}{\sqrt{1}-\sqrt{1}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}-\frac{1}{\sqrt{4}+\sqrt{5}}+....-\frac{1}{\sqrt{8}-\sqrt{2}}\)
a)\(\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+...+\frac{1}{\sqrt{1936}+\sqrt{1935}}=\)
\(\frac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}\)\(+\frac{\sqrt{4}-\sqrt{3}}{\left(\sqrt{4}-\sqrt{3}\right)\left(\sqrt{4}+\sqrt{3}\right)}+...\)\(+\frac{\sqrt{1936}-\sqrt{1935}}{\left(\sqrt{1936}-\sqrt{1935}\right)\left(\sqrt{1936}+\sqrt{1935}\right)}\)= \(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{1936}-\sqrt{1935}\)= \(-1-\sqrt{1935}\)
b)đề hơi sai bạn ạ mẫu thức số một bằng 0 còn đâu sửa lại đề đi nhé sau đó trục căn thức tương tự như mk làm nha
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cảm ơn bạn nha mik ghi dề sai đề đúng là như thế này nè\(\frac{1}{\sqrt{1}-\sqrt{2}}\) bạn giải giúp mik lun đi mik cảm ơn b nhìu lắm
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1,Rút gọn:
a, \(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+2}\)
b,\(\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}+\frac{1}{\sqrt{5}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{9}}\)
Nhân cả tử và mẫu với biểu thức liên hợp của mẫu (câu a mẫu cuối kì kì)
\(A=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\frac{1}{4}=\sqrt{3}-\frac{3}{4}\)
\(B=-\left(\sqrt{1}+\sqrt{2}-\sqrt{2}-\sqrt{3}+\sqrt{3}+\sqrt{4}-...+\sqrt{7}+\sqrt{8}-\sqrt{8}-\sqrt{9}\right)\)
\(B=-\left(\sqrt{1}-\sqrt{9}\right)=2\)
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Chứng minh rằng \(\frac{1}{3\sqrt[2]{2}}+\frac{1}{4\sqrt[3]{3}}+\frac{1}{5\sqrt[3]{4}}+.....+\frac{1}{1000\sqrt[3]{999}}< \frac{11}{5}\)
1-a,\(A=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
b,\(B=\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
Ta có :
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
Ta có:
\(\frac{1}{\sqrt{x}+\sqrt{x-1}}=\frac{\sqrt{x}-\sqrt{x-1}}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}-\sqrt{x-1}\right)}=\sqrt{x}-\sqrt{x-1}\)
Do đó:
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
\(\Leftrightarrow A=\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+\sqrt{3}-\sqrt{4}+...+\sqrt{n-1}+\sqrt{n}\)
\(\Leftrightarrow A=\sqrt{n}-1\left(dpcm\right)\)
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