\(\frac{3}{4}x\frac{8}{9}x...x\frac{120}{121}=\frac{1x3}{2x2}x\frac{2x4}{3x3}x...x\frac{10x12}{11x11}=\frac{1x3x2x4x...x10x12}{2x2x3x3x...x11x11}\)
\(\frac{\left(1x2x3x4x...x10\right)x\left(3x4x5x6x...x12\right)}{\left(2x3x4x5x...x11\right)x\left(2x3x4x5x...x11\right)}=\frac{12}{11x2}=\frac{6}{11}\)