\(\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{49.51}\)
\(=\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{49.51}\right)\)
\(=\frac{3}{2}.\left(1-\frac{1}{51}\right)\)
\(=\frac{25}{17}\)
Xin lỗi em mới học dạng này hả để anh làm lại cho hiểu nhé
\(\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{49.51}\)
\(=\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{49.51}\right)\)
\(=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\left(1-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\frac{50}{51}\)
\(=\frac{25}{17}\)
#)Giải :
Đặt \(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(A=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(A=\frac{3}{2}\times\frac{50}{51}\)
\(A=\frac{25}{17}\)
đề bài có vấn đề , mk sửa lại để chút nha : 3/1.3 +3/3.5+3/5.7+...+3/49.51
A=3/2(2/1.3+2/3.5+2/5.7+....+2/49.51)
=3/2(1-1/3+1/3-1/5+1/5-1/7+....+1/49.51)
=3/2(1-1/51)
=3/2.50/51
=25/17
Mấy bài này mik lên Bingbe nhé có câu hỏi tương tự đó
A=3/2(2/1.3+2/3.5+2/5.7+....+2/53.55)
=3/2(1-1/3+1/3-1/5+1/5-1/7+..../1/53-1/55)
=3/2(1-1/55)
=3/2.54/55
=81/55
~Hok tốt~
\(\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{49.51}\right)\)
Sửa đề vì sai đề:\(\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{49.51}\)
= \(\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{49.51}\right)\)
= \(\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)
= \(\frac{3}{2}.\left(1-\frac{1}{51}\right)\)
= \(\frac{3}{2}.\frac{50}{51}\)
= \(\frac{25}{17}\)
cảm ơn mọi người nhiều vì mình mới tìm hiểu giạng này mà hôm bữa ko đi học thêm nên ko biết làm
Sửa đề :\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
Đặt : \(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(\Leftrightarrow A=\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(\Leftrightarrow A=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(\Leftrightarrow A=\frac{3}{2}.\left[1+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+...+\left(\frac{1}{49}-\frac{1}{49}\right)-\frac{1}{51}\right]\)
\(\Leftrightarrow A=\frac{3}{2}.\left(1-\frac{1}{51}\right)\)
\(\Leftrightarrow A=\frac{3}{2}.\frac{50}{51}=\frac{25}{17}\)
~ Hok tốt ~
