1.
a,
\(\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{19\cdot21}\right)\cdot462-\left[2,04:\left(x+1,05\right)\right]:0,12=19\\ \left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\cdot462-\left[2,04:\left(x+1,05\right)\right]:0,12=19\\ \left(\dfrac{1}{11}-\dfrac{1}{21}\right)\cdot462-\left[2,04:\left(x+1,05\right)\right]:0,12=19\\ \dfrac{10}{231}\cdot462-\left[2,04:\left(x+1,05\right)\right]:0,12=19\\ 20-\left[2,04:\left(x+1,05\right)\right]:0,12=19\\ \left[2,04:\left(x+1,05\right)\right]:0,12=1\\ 2,04:\left(x+1,05\right)=0,12\\ x+1,05=17\\ x=15,95\)
b,
\(\dfrac{1}{24\cdot25}+\dfrac{1}{25\cdot26}+...+\dfrac{1}{29\cdot30}+x:\dfrac{1}{3}=-4\\ \dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{29}-\dfrac{1}{30}+x\cdot3=-4\\ \dfrac{1}{24}-\dfrac{1}{30}+x\cdot3=-4\\ \dfrac{1}{120}+x\cdot3=-4\\ 3x=\dfrac{-481}{120}\\ x=\dfrac{-481}{360}\)
2.
a,
\(\dfrac{15}{28}-\dfrac{186}{1116}-\dfrac{121}{462}+\dfrac{189}{198}\\ =\dfrac{15}{28}-\dfrac{1}{6}-\dfrac{11}{42}+\dfrac{21}{22}\\ =\dfrac{495}{924}-\dfrac{154}{924}-\dfrac{242}{924}+\dfrac{882}{924}\\ =\dfrac{495-154-242+882}{924}\\ =\dfrac{981}{924}\\ =\dfrac{327}{308}\)
b,
\(\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{1\cdot3}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{2\cdot4}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{3\cdot5}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{99\cdot101}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{2^2-1}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{3^2-1}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{4^2-1}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{100^2-1}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\)\(=\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{100^2}{99\cdot101}\\ =\dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{3\cdot3}{2\cdot4}\cdot\dfrac{4\cdot4}{3\cdot5}\cdot...\cdot\dfrac{100\cdot100}{99\cdot101}\\ =\dfrac{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot100\cdot100}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot99\cdot101}\\ =\dfrac{\left(2\cdot3\cdot4\cdot...\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot100\right)}{\left(1\cdot2\cdot3\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot101\right)}\\ =\dfrac{100\cdot2}{1\cdot101}\\ =\dfrac{200}{101}\)
