=2x2/1x3 x 3x3/2x4x4/3x5 x ... x 11x11/10x12

=2x2x3x3x4x4x...x11x11/1x3x2x4x3x5x...x10x12

= 2x3x4x...x11)x(2x3x4x...x11)/(1x2x3x4x...x11)/(1x2x3x...x10)x(2x3x4x...x11)

=11x2/1x12

=22/12

=11/6

4/3x9/8x16/15x...x121/120=\(\dfrac{2^2}{1\cdot3}\)x\(\dfrac{3^2}{2\cdot4}\)x\(\dfrac{4^2}{3\cdot5}\)x...x\(\dfrac{11^2}{10\cdot12}\)=\(\dfrac{2}{3}\)x\(\dfrac{11}{10}\)=\(\dfrac{11}{15}\)

4/3x9/8x16/15x...x121/120=\(\dfrac{2^2}{1\cdot3}\)x\(\dfrac{3^2}{2\cdot4}\)x...x\(\dfrac{11^2}{10\cdot12}\)=\(\dfrac{2}{3}\)x\(\dfrac{11}{10}\)=\(\dfrac{11}{15}\)

\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{120}{121}\)

\(=\frac{2.3.4......10}{2.3.4.....11}.\frac{3.4.5.....12}{2.3.4.....11}\)

\(=\frac{1}{11}.6=\frac{6}{11}\)

\(\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.\frac{25}{24}...\frac{121}{120}.\frac{144}{143}\)

\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}...\frac{11.11}{10.12}.\frac{12.12}{11.13}\)

\(=\frac{2.2.3.3.4.4.5.5...11.11.12.12}{1.3.2.4.3.5.4.6...10.12.11.13}\)

\(=\frac{\left(2.3.4.5...11.12\right).\left(2.3.4.5...11.12\right)}{\left(1.2.3.4...10.11\right).\left(3.4.5.6...12.13\right)}\)

\(=\frac{12.2}{1.13}\)

\(=\frac{24}{13}\)

  ( Dấu \(.\)là dấu \(\times\)nha )

Ta có:\(B=\frac{1.3}{2.2}x\frac{2.4}{3.3}x\frac{3.5}{4.4}x...x\frac{10.12}{11.11}\)

                          \(=\frac{1.2.3.4.5.6.7.8.9.10}{2.3.4.5.6.7.8.9.10.11}x\frac{3.4.5.6.7.8.9.10.11.12}{2.3.4.5.6.7.8.9.10.11}\)

                           \(=\frac{1}{11}x6\)

                             \(=\frac{6}{11}\)

             Ai thấy đúng k nha.

mk cũng ko biết bn à

\(\frac{5-\frac{5}{3}+\frac{5}{9}-\frac{5}{27}}{8-\frac{8}{3}+\frac{8}{9}-\frac{8}{27}}:\frac{15-\frac{15}{11}+\frac{15}{121}}{16-\frac{16}{11}+\frac{16}{121}}\)

\(=\frac{5\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}{8\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}:\frac{15\left(1-\frac{1}{11}+\frac{1}{121}\right)}{16\left(1-\frac{1}{11}+\frac{1}{121}\right)}\)

\(=\frac{5}{8}:\frac{15}{16}\)

\(=\frac{2}{3}\)

2/3

Tk mk nha