Ta có:
\(\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{49.51}\)
\(=\dfrac{3}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)
\(=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(=\dfrac{3}{2}\left(1-\dfrac{1}{51}\right)\)
\(=\dfrac{3}{2}.\dfrac{50}{51}=\dfrac{25}{17}\)
\(\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{49.51}\)
=\(\dfrac{3}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)
=\(\dfrac{3}{2}\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\right)\)
\(=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(=\dfrac{3}{2}\left(1-\dfrac{1}{51}\right)\)
=\(\dfrac{3}{2}.\dfrac{50}{51}\)=\(\dfrac{25}{17}\)
~ chúc bn học tốt~
Viết các tích sau thành dạng lũy thừa của một số nguyên :
a) (-8).(-3)3.(+125)
b) 27.(-2)3.(-7).(+49)
