\(\frac{\frac{3}{7}+\frac{3}{103}+\frac{3}{17}}{\frac{5}{7}+\frac{5}{103}+\frac{5}{17}}=?\)
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6 tháng 3 2021
Xét biểu thức phụ :
1
(2n+3)√2n+1+(2n+1)√2n+3 =
1
√2n+1.√2n+3(√2n+1+√2n+3)
=
√2n+3−√2n+1
√2n+1.√2n+3[(2n+3)−(2n+1)]
=
√2n+3−√2n+1
2√2n+1.√2n+3 =
1
2 (
1
√2n+1 −
1
√2n+3 )với
n≥1
Áp dụng :
S=
1
3√1+1√3 +
1
3√5+5√3 +
1
5√7+7√5 +...+
1
101√103+103√101
=
1
2 (
1
√1 −
1
√3 )+
1
2 (
1
√3 −
1
√5 )+
1
2 (
1
√5 −
1
√7 )+...+
1
2 (
1
√101 −
1
√103 )
=
1
2 (1−
1
√3 +
1
√3 −
1
√5 +
1
√5 −
1
√7 +...+
1
√101 −
1
√103 )
=
1
2 (1−
1
√103 )
6 tháng 8 2016
Xét biểu thức phụ : \(\frac{1}{\left(2n+3\right)\sqrt{2n+1}+\left(2n+1\right)\sqrt{2n+3}}=\frac{1}{\sqrt{2n+1}.\sqrt{2n+3}\left(\sqrt{2n+1}+\sqrt{2n+3}\right)}\)
\(=\frac{\sqrt{2n+3}-\sqrt{2n+1}}{\sqrt{2n+1}.\sqrt{2n+3}\left[\left(2n+3\right)-\left(2n+1\right)\right]}\)
\(=\frac{\sqrt{2n+3}-\sqrt{2n+1}}{2\sqrt{2n+1}.\sqrt{2n+3}}=\frac{1}{2}\left(\frac{1}{\sqrt{2n+1}}-\frac{1}{\sqrt{2n+3}}\right)\)với \(n\ge1\)
Áp dụng : \(S=\frac{1}{3\sqrt{1}+1\sqrt{3}}+\frac{1}{3\sqrt{5}+5\sqrt{3}}+\frac{1}{5\sqrt{7}+7\sqrt{5}}+...+\frac{1}{101\sqrt{103}+103\sqrt{101}}\)
\(=\frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{3}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}\right)+...+\frac{1}{2}\left(\frac{1}{\sqrt{101}}-\frac{1}{\sqrt{103}}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}+...+\frac{1}{\sqrt{101}}-\frac{1}{\sqrt{103}}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{103}}\right)\)
6 tháng 3 2021
Xét biểu thức phụ :
1
(2n+3)√2n+1+(2n+1)√2n+3 =
1
√2n+1.√2n+3(√2n+1+√2n+3)
=
√2n+3−√2n+1
√2n+1.√2n+3[(2n+3)−(2n+1)]
=
√2n+3−√2n+1
2√2n+1.√2n+3 =
1
2 (
1
√2n+1 −
1
√2n+3 )với
n≥1
Áp dụng :
S=
1
3√1+1√3 +
1
3√5+5√3 +
1
5√7+7√5 +...+
1
101√103+103√101
=
1
2 (
1
√1 −
1
√3 )+
1
2 (
1
√3 −
1
√5 )+
1
2 (
1
√5 −
1
√7 )+...+
1
2 (
1
√101 −
1
√103 )
=
1
2 (1−
1
√3 +
1
√3 −
1
√5 +
1
√5 −
1
√7 +...+
1
√101 −
1
√103 )
=
1
2 (1−
1
√103 )
17 tháng 9 2015
\(A=\frac{3.\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}{5.\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{1.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}{\left(-7\right).\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
\(A=\frac{3}{5}+\frac{-1}{7}\)
\(A=\frac{21}{35}+\frac{-5}{35}\)
\(A=\frac{16}{35}\)
MH
17 tháng 9 2015
\(A=\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
\(=\frac{3.\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}{5.\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-\frac{7}{2}+\frac{7}{3}-\frac{7}{4}+\frac{7}{5}}\)
\(=\frac{3}{5}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-7.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
=3/5+(1/-7)
=3/5-1/7
=16/35
14 tháng 4 2019
Cac ban giup mik voi, mik k cho! ( 10 k cho 3 nguoi dau tien tra loi cau hoi cua mik)
5 tháng 7 2016
\(A=\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{17}-\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}-\frac{1}{37}\right)}+\frac{1.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
\(=\frac{3}{5}+\frac{-1}{7}\)
\(=\frac{21}{35}-\frac{5}{35}\)
\(=\frac{16}{35}\)
5 tháng 7 2016
\(A=\frac{3.\left(\frac{1}{7}-\frac{1}{17}-\frac{1}{37}\right)}{5.\left(\frac{1}{7}-\frac{1}{17}-\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{7.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
\(A=\frac{3}{5}+\frac{1}{7}=\frac{21}{35}+\frac{5}{35}=\frac{26}{35}\)
5 tháng 7 2019
A=\(\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
\(=\frac{3}{5}+\frac{1}{-7}=\frac{3}{5}-\frac{1}{7}\)
\(=\frac{21}{35}-\frac{5}{35}=\frac{16}{35}\)
29 tháng 8 2015
\(A=\frac{3\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}=\frac{3}{5}+\frac{1}{7}=\frac{26}{35}\)
29 tháng 8 2015
\(A=\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}=\frac{3.\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}{5.\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{-1}{5}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{-7}{2}+\frac{7}{5}-\frac{7}{4}+\frac{7}{3}}\)
\(A=\frac{3}{5}+\frac{\frac{1}{2}-\frac{1}{5}-\frac{1}{3}+\frac{1}{4}}{\frac{7}{5}-\frac{7}{2}-\frac{7}{4}+\frac{7}{3}}=\frac{3}{5}+\frac{\frac{1}{20}}{7.\frac{-13}{60}}=\frac{3}{5}+\frac{-3}{91}=\frac{258}{455}\)
23 tháng 3 2019
\(\Leftrightarrow\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\Rightarrow x\in\left(-2;-1;0;1;2\right)\)
\(\Leftrightarrow\frac{-1}{24}\le\frac{x}{24}\le\frac{5}{24}\Rightarrow x\in\left(-1;0;1;2;3;4;5\right)\)
2 câu sau tự làm nha
23 tháng 3 2019
\(-\frac{5}{17}+\frac{3}{17}\le\frac{x}{17}\le\frac{13}{17}+-\frac{11}{17}\)
\(\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\)
=> \(x\in\left\{-2;-1;0;1;2\right\}\)