\(\frac{3}{10}+\frac{3}{90}+\frac{3}{234}+....+\frac{3}{1050}\)
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\(\frac{3}{10}+\frac{3}{90}+\frac{3}{324}+...+\frac{3}{1050}\)
\(\frac{3}{10}+\frac{3}{90}+\frac{3}{324}+...+\frac{1}{1050}\)
Tính hợp lí:
a) \(A=\left(\frac{3}{4}-\frac{4}{7}-\frac{5}{6}\right):\left(-\frac{3}{10}+\frac{4}{14}-\frac{-5}{12}\right)\left(\frac{1}{12}-\frac{7}{6}\right)\)
b) 123.(-234+4356-2312)+234.(123-2312)-2312.(-234-123)
a: \(A=\dfrac{63-48-70}{84}:\dfrac{-3\cdot84+4\cdot60+5\cdot70}{840}\cdot\dfrac{1-14}{12}\)
\(=\dfrac{-55}{84}\cdot\dfrac{840}{338}\cdot\dfrac{-13}{12}=\dfrac{55}{1}\cdot\dfrac{10}{338}\cdot\dfrac{13}{12}=\dfrac{275}{156}\)
b: \(=-234\cdot123+123\cdot4356-123\cdot2312+234\cdot123-234\cdot2312+2312\cdot234+2312\cdot123\)
\(=123\cdot4356-123\cdot2312+123\cdot2312=123\cdot4356=535788\)
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Chứng minh rằng \(\frac{1}{3^1}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{90}}-\frac{1}{3^{100}}<\frac{3}{10}\)
\(\frac{3}{10}+\frac{3}{90}+\frac{3}{324}+...+\frac{3}{1050}\) Ai trả lời đúng dầu tiên mình sẽ tick cho nhá . Gấp . MỌI NGƯỜI GIÚP MÌNH VỚI NHÉ .
A=\(\frac{10-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{10}{18}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{90}}\)
\(\frac{3}{30}+\frac{3}{42}+\frac{3}{56}+\frac{3}{72}+\frac{3}{90}+\frac{3}{110}+\frac{3}{132}=\frac{1}{6}X\)
\(\frac{3}{5.6}+\frac{3}{6.7}+......+\frac{3}{11.12}=\frac{1}{6}X\)
\(\Rightarrow3.\left(\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{11.12}\right)=\frac{1}{6}X\)
\(\Rightarrow3.\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+.....+\frac{1}{11}-\frac{1}{12}\right)=\frac{1}{6}X\)
\(\Rightarrow3.\left(\frac{1}{5}-\frac{1}{12}\right)=\frac{1}{6}X\)
\(\Rightarrow3.\frac{7}{60}=\frac{1}{6}X\)
\(\Rightarrow\frac{21}{60}=\frac{1}{6}X\)
\(\Rightarrow X=\frac{21}{60}\div\frac{1}{6}=\frac{21}{10}\)
Vậy \(X=\frac{21}{10}\)
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\(\frac{x-1}{99}+\frac{x-3}{97}=\frac{x-4}{96}+\frac{x-10}{90}\)
B=\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+10}\)
C=\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
Ta có công thức : \(1+2+3+...+n=\frac{n.\left(n+1\right)}{2}\)
\(\Rightarrow B=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+10}\)
\(=\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+3\right).3}{2}}+...+\frac{1}{\frac{\left(1+10\right)10}{2}}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{10.11}\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{11}\right)=2.\frac{9}{22}=\frac{9}{11}\)
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