\(\frac{3}{2}\)x \(\frac{3}{4}\)x \(\frac{4}{5}\): \(\frac{1}{5}\)= \(\frac{9}{8}\)x \(\frac{4}{5}\): \(\frac{1}{5}\)= \(\frac{9}{10}\): \(\frac{1}{5}\)= \(\frac{9}{10}\)x \(\frac{5}{1}\)= \(\frac{9}{2}\)

bằng 36 phần 8 thì phải

\(\dfrac{2}{3}\times\left(x+\dfrac{4}{5}\right)=\dfrac{-1}{3}\\ x+\dfrac{4}{5}=\dfrac{-1}{3}:\dfrac{2}{3}\\ x+\dfrac{4}{5}=\dfrac{-1}{3}\times\dfrac{3}{2}\\ x+\dfrac{4}{5}=\dfrac{-1}{2}\\ x=\dfrac{-1}{2}-\dfrac{4}{5}\\ x=\dfrac{-5}{10}-\dfrac{8}{10}\\ x=\dfrac{-13}{10}\)

\(\dfrac{2}{3}.\left(x+\dfrac{4}{5}\right)=-\dfrac{1}{3}\)

    \(\left(x+\dfrac{4}{5}\right)=\left(-\dfrac{1}{3}:\dfrac{2}{3}\right)\)

    \(\left(x+\dfrac{4}{5}\right)=\left(-\dfrac{1}{3}.\dfrac{3}{2}\right)=-\dfrac{1}{2}\)

      \(x\)            \(=\left(-\dfrac{1}{2}\right)-\dfrac{4}{5}\)

      \(x\)            \(=\left(-\dfrac{1}{2}\right)+\left(-\dfrac{4}{5}\right)\)

      \(x\)            \(=-\dfrac{13}{10}\)

\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{100^2}-1\right)=\frac{-3}{2^2}.\frac{-8}{3^2}...\frac{-9999}{100^2}\)

\(=-\frac{3.8...9999}{2^2.3^2...100^2}=-\frac{1.3.2.4...99.101}{2.2.3.3...100.100}=-\frac{\left(1.2....99\right).\left(3.4...101\right)}{\left(2.3...100\right).\left(2.3...100\right)}=-\frac{1.101}{100.2}=-\frac{101}{200}\)

\(< -\frac{100}{200}=\frac{1}{2}=B\)

=> A < B

15/30+2/5=1/2+2/5=9/10

15/30 + 2/5 = 1/2 +2/5 = 9/10

15/30+2/5=1/2+2/5=9/10

Đặt: \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2011.2013}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2011.2013}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2013}\right)\)

\(=\frac{1}{2}.\frac{2012}{2013}\)

\(=\frac{1006}{2013}\)

51/30

24/65

chưa xong 

\(\dfrac{3}{4}.\dfrac{2}{3}-\dfrac{1}{6}.\dfrac{3}{4}-\dfrac{3}{4}.\dfrac{1}{2}\) 

= \(\dfrac{3}{4}.\left(\dfrac{2}{3}-\dfrac{1}{6}-\dfrac{1}{2}\right)\) 

= \(\dfrac{3}{4}.0\) 

= 0

\(\dfrac{3}{4}\cdot\dfrac{2}{3}-\dfrac{1}{6}\cdot\dfrac{3}{4}-\dfrac{3}{4}\cdot\dfrac{1}{2}\)

\(=\dfrac{3}{4}\left(\dfrac{2}{3}-\dfrac{1}{6}-\dfrac{1}{2}\right)\)

\(=\dfrac{3}{4}\cdot\left(\dfrac{4}{6}-\dfrac{1}{6}-\dfrac{3}{6}\right)\)

=0

a: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{100\cdot101}\)

=1-1/2+1/2-1/3+...+1/100-1/101

=1-1/101=100/101

b: \(A=1+\dfrac{1}{2}+1+\dfrac{1}{6}+1+\dfrac{1}{12}+...+1+\dfrac{1}{10100}\)

\(=100+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{100}-\dfrac{1}{101}\right)\)

\(=101-\dfrac{1}{101}< 101\)