\(\frac{2,3}{7}+\left(\frac{2}{9}-\frac{10}{7}\right)-\frac{5}{3}:\frac{1}{9}\)

\(\frac{2,3}{7}-\frac{76}{63}-15\)

\(-\frac{55,3}{63}-15\)

\(-15\frac{55,3}{63}\)

a: \(x-\frac25+\frac{7}{12}=\frac12\cdot\frac34\cdot\frac{-9}{16}\)

=>\(x-\frac{24}{60}+\frac{35}{60}=\frac38\cdot\frac{-9}{16}=\frac{-27}{128}\)

=>\(x=-\frac{27}{128}-\frac{11}{60}=\frac{-27\cdot15-11\cdot32}{1920}=-\frac{757}{1920}\)

b: \(\frac49x-\frac12=-\frac59\)

=>\(\frac49x=-\frac59+\frac12=\frac{-10+9}{18}=\frac{-1}{18}\)

=>\(x=-\frac{1}{18}:\frac49=-\frac{1}{18}\cdot\frac94=\frac{-1}{2\cdot4}=-\frac18\)

c: \(-\frac23x+\frac58=\frac{7}{12}\)

=>\(-\frac23x=\frac{7}{12}-\frac58=\frac{14}{24}-\frac{15}{24}=-\frac{1}{24}\)

=>\(x=\frac{1}{24}:\frac23=\frac{1}{24}\cdot\frac32=\frac{1}{8\cdot2}=\frac{1}{16}\)

d: \(\frac25x:\frac73\cdot\frac18=\frac{29}{15}\)

=>\(x\cdot\frac25\cdot\frac37\cdot\frac18=\frac{29}{15}\)

=>\(x\cdot\frac{3}{4\cdot5\cdot7}=\frac{29}{15}\)

=>\(x=\frac{29}{15}:\frac{3}{140}=\frac{29}{15}\cdot\frac{140}{3}=\frac{29\cdot28}{3\cdot3}=\frac{812}{9}\)

- \(\dfrac{1}{9}\) \(\times\) ( - \(\dfrac{3}{5}\)) + \(\dfrac{5}{-6}\) \(\times\) ( \(-\dfrac{3}{5}\)) - \(\dfrac{7}{2}\) \(\times\) \(\dfrac{3}{5}\)

= \(\dfrac{1}{9}\) \(\times\) \(\dfrac{3}{5}\) + \(\dfrac{5}{6}\) \(\times\) \(\dfrac{3}{5}\) - \(\dfrac{7}{2}\) \(\times\) \(\dfrac{3}{5}\)

= \(\dfrac{3}{5}\) \(\times\) ( \(\dfrac{1}{9}\) + \(\dfrac{5}{6}\) - \(\dfrac{7}{2}\))

= \(\dfrac{3}{5}\) \(\times\) ( \(\dfrac{2}{18}\) + \(\dfrac{15}{18}\) - \(\dfrac{63}{18}\))

= \(\dfrac{3}{5}\) \(\times\) (- \(\dfrac{23}{9}\))

= - \(\dfrac{23}{15}\)

Ta có :

\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)

\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)

\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)

\(=1-\frac{1}{10^2}< 1\)

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)

\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{19}{81.100}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{81}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}< 1\left(dpcm\right)\) 

CS AI XEM S** KO

\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)

\(=\left(\frac{1}{1^2}-\frac{1}{2^2}\right)+\left(\frac{1}{2^2}-\frac{1}{3^2}\right)+\left(\frac{1}{3^2}-\frac{1}{4^2}\right)+...+\left(\frac{1}{9^2}-\frac{1}{10^2}\right)\)

\(=\frac{1}{1}-\frac{1}{10^2}\)

\(=1-\frac{1}{100}<1\)

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)

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)

\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{19}{81.100}\)\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{81}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}< 1\)