\(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\dfrac{15}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow x=15\)

Vậy x = 15

a: \(=\dfrac{2^{19}\cdot3^9+3^9\cdot5\cdot2^{18}}{2^{19}\cdot3^9+2^{10}}\)

\(=\dfrac{3^9\cdot2^{18}\cdot\left(2+5\right)}{2^{10}\cdot\left(2^9\cdot3^9+1\right)}=\dfrac{3^9\cdot7\cdot2^8}{6^9+1}\)

b: \(=\dfrac{\dfrac{-1}{8}-\dfrac{27}{64}\cdot4}{-2+\dfrac{9}{16}-\dfrac{3}{8}}=\dfrac{-29}{16}:\dfrac{-29}{16}=1\)

1: \(\left(-\frac{25}{13}\right)+\left(-\frac{19}{17}\right)+\frac{12}{13}+\left(-\frac{25}{17}\right)\)

\(=\left(-\frac{25}{13}+\frac{12}{13}\right)+\left(-\frac{19}{17}-\frac{15}{17}\right)\)

\(=-\frac{13}{13}-\frac{34}{17}=-1-2=-3\)

2: \(\frac12-\left(-\frac13\right)+\frac{1}{23}+\frac16\)

\(=\left(\frac12+\frac13+\frac16\right)+\frac{1}{23}\)

\(=1+\frac{1}{23}=\frac{24}{23}\)

3: \(\left(-\frac37\right)\cdot\frac{5}{11}+\left(-\frac{5}{14}\right)\cdot\frac{5}{11}\)

\(=\frac{5}{11}\left(-\frac37-\frac{5}{14}\right)\)

\(=\frac{5}{11}\cdot\left(-\frac{6}{14}-\frac{5}{14}\right)=\frac{5}{11}\cdot\frac{-11}{14}=\frac{-5}{14}\)

4: \(\left(-\frac{5}{11}\right)\cdot\frac{7}{15}\cdot\frac{11}{-5}\cdot\left(-30\right)=-30\cdot\frac{7}{15}=-14\)

5: \(\left(-\frac59\right)\cdot\frac{3}{11}+\left(-\frac{13}{18}\right)\cdot\frac{3}{11}\)

\(=\frac{3}{11}\left(-\frac59-\frac{13}{18}\right)\)

\(=\frac{3}{11}\cdot\frac{-23}{18}=\frac{-23\cdot1}{11\cdot6}=-\frac{23}{66}\)

6: \(2\frac{2}{15}\cdot\frac{9}{17}\cdot\frac{3}{32}:\left(-\frac{3}{17}\right)\)

\(=\frac{32}{15}\cdot\frac{3}{32}\cdot\frac{9}{17}\cdot\frac{-17}{3}=\frac{3}{15}\cdot\frac{-9}{3}=\frac{-9}{15}=-\frac35\)

7: \(\left(-\frac34+\frac25\right):\frac37+\left(\frac35-\frac14\right):\frac37\)

\(=\left(-\frac34+\frac25+\frac35-\frac14\right):\frac37\)

\(=\left(-1+1\right):\frac37=0\)

8: \(\left(-\frac13\right)\cdot\left(-\frac{15}{19}\right)\cdot\frac{38}{45}=\frac{15}{45\cdot3}\cdot\frac{38}{19}=2\cdot\frac{1}{3\cdot3}=\frac29\)

9: \(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots+\frac{1}{19\cdot20}\)

\(=\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{19}-\frac{1}{20}\)

\(=\frac12-\frac{1}{20}=\frac{9}{20}\)

10: \(\frac{1}{9\cdot10}-\frac{1}{8\cdot9}-\frac{1}{7\cdot8}-\cdots-\frac{1}{2\cdot3}-\frac{1}{1\cdot2}\)

\(=\frac{1}{9\cdot10}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{8\cdot9}\right)\)

\(=\frac19-\frac{1}{10}-\left(1-\frac12+\frac12-\frac13+\cdots+\frac18-\frac19\right)\)

\(=\frac19-\frac{1}{10}-\left(1-\frac19\right)=\frac19-\frac{1}{10}-\frac89=-\frac79-\frac{1}{10}=\frac{-79}{90}\)

1,

\(A=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2018}-1\right)\\ A=\left(-\dfrac{1}{2}\right)\cdot\left(-\dfrac{2}{3}\right)\cdot...\cdot\left(-\dfrac{2017}{2018}\right)\\ =-\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2017}{2018}\right)\\ =-\dfrac{1}{2018}\)

\(\frac{\left(\frac{2}{3}\right)^3.\left(\frac{-3}{4}\right)^2.\left(-1\right)^5}{\left(\frac{2}{5}\right)^2.\left(\frac{-5}{12}\right)^3}=\frac{\frac{2^3}{3^3}.\frac{3^2}{4^2}.\left(-1\right)}{\frac{2^2}{5^2}.\frac{\left(-5\right)^3}{12^3}}=\)\(\frac{\frac{1}{6}.\left(-1\right)}{\frac{2^2}{5^2}.\frac{5^3}{2^6.3^3}.\left(-1\right)}=\frac{\frac{1}{2.3}}{\frac{5}{2^4.3^3}}=\frac{2^3.3^2}{5}=\frac{72}{5}\)

a: \(=\dfrac{3}{8}\left(27+\dfrac{1}{5}-51-\dfrac{1}{5}\right)+19\)

\(=-24\cdot\dfrac{3}{8}+19=-9+19=10\)

b: \(=\left(35+\dfrac{1}{6}-46-\dfrac{1}{6}\right):\left(\dfrac{-4}{5}\right)\)

\(=\dfrac{-11\cdot5}{-4}=\dfrac{55}{4}\)

c: \(=\left(\dfrac{-15+8}{20}\right):\left[\dfrac{3}{7}+\dfrac{7}{3}\cdot\dfrac{12-5}{20}\right]\)

\(=\dfrac{-7}{20}:\left(\dfrac{3}{7}+\dfrac{49}{60}\right)\)

\(=-\dfrac{147}{523}\)