Question

[Nguyễn Hữu Cường]

Vì viết phân số 0 được nên mình viết đở /

a. 3/1×3+3/3×5+3/5×7+....+3/49×51=?

Answer 1

[Hoang Hung Quan]

Ta có:

\(\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{49.51}\)

\(=\dfrac{3}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)

\(=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)

\(=\dfrac{3}{2}\left(1-\dfrac{1}{51}\right)\)

\(=\dfrac{3}{2}.\dfrac{50}{51}=\dfrac{25}{17}\)

Answer 2

[Phạm Thanh Hằng]

\(\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{49.51}\)

=\(\dfrac{3}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)

=\(\dfrac{3}{2}\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\right)\)

\(=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)

\(=\dfrac{3}{2}\left(1-\dfrac{1}{51}\right)\)

=\(\dfrac{3}{2}.\dfrac{50}{51}\)=\(\dfrac{25}{17}\)

~ chúc bn học tốt~