Bài 1:
a: \(\left(-3\right)\cdot\dfrac{-7}{12}=3\cdot\dfrac{7}{12}=\dfrac{21}{12}=\dfrac{7}{4}\)
b: \(-\dfrac{3}{9}-\dfrac{8}{12}=-\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{3}{3}=-1\)
Bài 4:
\(\dfrac{4}{1\cdot3}+\dfrac{4}{3\cdot5}+...+\dfrac{4}{99\cdot101}\)
\(=2\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)
\(=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=2\left(1-\dfrac{1}{101}\right)=2\cdot\dfrac{100}{101}=\dfrac{200}{101}\)
Bài 5:
\(A=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{16}\left(1+2+...+16\right)\)
\(=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{16}\cdot\dfrac{16\cdot17}{2}\)
\(=\dfrac{2}{2}+\dfrac{3}{2}+...+\dfrac{17}{2}\)
\(=\dfrac{\left(2+3+...+17\right)}{2}=\dfrac{\dfrac{16\left(17+2\right)}{2}}{2}=\dfrac{8\left(17+2\right)}{2}=4\cdot19=76\)
Bài 2
a) \(\left(\dfrac{3}{5}\right).\left(-\dfrac{7}{12}\right)=-\dfrac{21}{60}=-\dfrac{7}{20}\)
b) \(\dfrac{3}{7}-\dfrac{8}{14}+1\)
\(=\dfrac{6}{14}-\dfrac{8}{14}+\dfrac{14}{14}\)
\(=\dfrac{12}{14}=\dfrac{6}{7}\)