\(a,\dfrac{-315}{540}=\dfrac{45.\left(-7\right)}{45.12}=\dfrac{-7}{12}\)
\(f,\dfrac{2929-101}{2.1919+404}=\dfrac{101\left(29-1\right)}{101\left(19.2+4\right)}=\dfrac{28}{42}=\dfrac{2}{3}\)
\(c,\dfrac{25.13}{26.35}=\dfrac{25.13}{13.70}=\dfrac{25}{70}=\dfrac{5}{14}\)
\(g,\dfrac{-1997.1996+1}{\left(-1995\right)\left(-1997\right)+1996}\)
\(=\dfrac{-1997.\left(1995+1\right)+1}{1995.1997+1996}\)
\(=\dfrac{-1997.1995-1997+1}{1997.1995+1996}\)
\(=\dfrac{-\left(1997.1995+1996\right)}{1997.1995+1996}=-1\)
Em cần bài nào đó em?
Bài 10:
a: \(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)
b: \(=\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{5}{2}\cdot\dfrac{100}{101}=\dfrac{500}{202}=\dfrac{250}{101}\)
