2.3/7+(2/9-1 3/7)-5/3:1/9
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\(\frac{2,3}{7}+\left(\frac{2}{9}-\frac{10}{7}\right)-\frac{5}{3}:\frac{1}{9}\)
\(\frac{2,3}{7}-\frac{76}{63}-15\)
\(-\frac{55,3}{63}-15\)
\(-15\frac{55,3}{63}\)
- \(\dfrac{1}{9}\) \(\times\) ( - \(\dfrac{3}{5}\)) + \(\dfrac{5}{-6}\) \(\times\) ( \(-\dfrac{3}{5}\)) - \(\dfrac{7}{2}\) \(\times\) \(\dfrac{3}{5}\)
= \(\dfrac{1}{9}\) \(\times\) \(\dfrac{3}{5}\) + \(\dfrac{5}{6}\) \(\times\) \(\dfrac{3}{5}\) - \(\dfrac{7}{2}\) \(\times\) \(\dfrac{3}{5}\)
= \(\dfrac{3}{5}\) \(\times\) ( \(\dfrac{1}{9}\) + \(\dfrac{5}{6}\) - \(\dfrac{7}{2}\))
= \(\dfrac{3}{5}\) \(\times\) ( \(\dfrac{2}{18}\) + \(\dfrac{15}{18}\) - \(\dfrac{63}{18}\))
= \(\dfrac{3}{5}\) \(\times\) (- \(\dfrac{23}{9}\))
= - \(\dfrac{23}{15}\)
Ta có :
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)
\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)
\(=1-\frac{1}{10^2}< 1\)
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{19}{81.100}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{81}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}< 1\left(dpcm\right)\)
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\left(\frac{1}{1^2}-\frac{1}{2^2}\right)+\left(\frac{1}{2^2}-\frac{1}{3^2}\right)+\left(\frac{1}{3^2}-\frac{1}{4^2}\right)+...+\left(\frac{1}{9^2}-\frac{1}{10^2}\right)\)
\(=\frac{1}{1}-\frac{1}{10^2}\)
\(=1-\frac{1}{100}<1\)
Vậy _____________________
=3/1.4+5/4.9+7/9.16+......+19/81.100
=(1/1-1/4)+(1/4-1/9)+........+(1/81-1/100)
=1-1/100
=99/100<1(đpcm)
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{19}{81.100}\)\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{81}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}< 1\)