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\(2014:\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1\frac{2}{5}-\frac{7}{9}+\frac{7}{11}}\cdot\frac{1\frac{1}{6}+0,875-0,7}{\frac{1}{3}+0,25-\frac{1}{5}}\right)\)
\(=2014:\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}\cdot\frac{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}}{\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}\right)\)
\(=2014:\left(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}\cdot\frac{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}}{\frac{2}{6}+\frac{2}{8}-\frac{2}{10}}\right)\)
\(=2014:\left(\frac{2}{7}\cdot\frac{7\left(\frac{1}{6}+\frac{1}{8}-\frac{1}{10}\right)}{2\left(\frac{1}{6}+\frac{1}{8}-\frac{1}{10}\right)}\right)\)
\(=2014:\left(\frac{2}{7}\cdot\frac{7}{2}\right)=2014\)
Ta có : \(1+2=\frac{2.3}{2}\) , \(1+2+3=\frac{3.4}{2}\) ,
\(1+2+3+4=\frac{4.5}{2}\) , ......... , \(1+2+3+4+....+2014=\frac{2014.2015}{2}\)
Suy ra : \(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2014.2015}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2014.2015}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(2\left(\frac{1}{2}-\frac{1}{2015}\right)\)
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2014}\)
\(A=\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+2014\right).2014:2}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2014.2015}\)
\(A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2014.2015}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{2015}\right)\)
\(A=2.\frac{1}{2}-2.\frac{1}{2015}\)
\(A=1-\frac{2}{2015}\)
\(A=\frac{2013}{2015}\)
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2014}\)
\(A=\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+2014\right).2014:2}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2014.2015}\)
\(A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{2015}\right)\)
\(A=2.\frac{1}{2}-2.\frac{1}{2015}\)
\(A=1-\frac{2}{2015}=\frac{2013}{2015}\)
A=2/6+2/12+....+2/4054182
A=2/2.3+2/3.4+...+2/2013.2014
A= (1-2/2014) : 2=503/1007
a: \(A=\left(0+1\right):\left(\dfrac{2}{3}+\dfrac{7}{6}-\dfrac{1}{6}\right)=1:\dfrac{5}{3}=\dfrac{3}{5}\)
b: \(B=\left[0.8\cdot15\right]\cdot\left[1.25\cdot\dfrac{19}{3}\right]+31.64=15\cdot\dfrac{95}{12}+31.64=150.39\)
Bạn xem lại đề câu a) cho rõ lại
Câu b) Tại x=2013 thì B=x2013-(x+1)x2012+(x+1)x2011-(x+1)x2010+...-(x+1)x2+(x+1)x-1
= x2013-x2013-x2012+x2012+x2011-x2011-x2010+..-x3 - x2+x2+x-1
= x-1 = 2012
a: \(A=\dfrac{1}{9}:\dfrac{1}{9}:\left(\dfrac{10+7}{15}:\dfrac{12-5}{30}\right)\)
\(=1:\left(\dfrac{17}{15}\cdot\dfrac{30}{7}\right)=1:\dfrac{34}{7}=\dfrac{7}{34}\)
b: \(=\left(5.6+0.64\right)\cdot1.25\cdot\dfrac{19}{3}+31.64\)
\(=\dfrac{39}{5}\cdot\dfrac{19}{3}+\dfrac{791}{25}=\dfrac{2026}{25}\)
\(\approx0,4\)