Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(M=\frac{3}{5.7}+\frac{3}{7.9}+....+\frac{3}{61.63}\)
\(2M=2.\left(\frac{3}{5.7}+\frac{3}{7.9}+.....+\frac{3}{61.63}\right)\)
\(2M=3.\left(\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{61.63}\right)\)
\(2M=3.\left(\frac{1}{5}-\frac{1}{63}\right)\)
\(2M=\frac{3.58}{315}=\frac{58}{105}\)
\(M=\frac{58}{105}.\frac{1}{2}=\frac{29}{105}\)
Ta có thể vt gọn thành :
M = \(\frac{3}{2}\).( \(\frac{1}{5}\)\(-\)\(\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\)... \(+\frac{1}{61}-\frac{1}{63}\))
M = \(\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{63}\right)\)
M = \(\frac{3}{2}.\frac{58}{315}\)
M = \(\frac{29}{105}\)
\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(=\frac{2}{3}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{51}\right)\)
\(=\frac{2}{3}.\frac{50}{51}=\frac{20}{51}\)
Ủng hộ mk nha !!! ^_^
\(\frac{3}{1.3}\)+ \(\frac{3}{3.5}\)+ \(\frac{3}{5.7}\)+...+ \(\frac{3}{49.51}\)
= \(\frac{3}{2}\)( \(\frac{2}{1.3}\)+ \(\frac{2}{3.5}\)+ \(\frac{2}{5.7}\)+...+ \(\frac{2}{49.51}\))
= \(\frac{3}{2}\)( \(\frac{1}{1}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+...+ \(\frac{1}{49}\)- \(\frac{1}{51}\))
= \(\frac{3}{2}\)( 1- \(\frac{1}{51}\))
= \(\frac{3}{2}\). \(\frac{50}{51}\)
= \(\frac{25}{17}\).
= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 +....... + 1/97 - 1/99
= 1- 1/99
= 98/99
=3.2/1.3.2+3.2/3.5.2+...+3.2/49.51
=3/2.(2/1.3+2/3.5+2/5.7+...+2/49.51)
=3/2.(1-1/3+1/3-1/5+...+1/49-1/51)
=3/2.(1-1/51)
=3/2.50/51
=25/17
CHÚC BẠN HỌC GIỎI
K MÌNH NHÉ
A=3/2(2/3.5+2/5.7+...+2/61.63)
=3/2(1/3-1/5+1/5-1/7+...+1/61-1/63)= 3/2(1/3-1/63)=3/2 x 20/63=10/21
Đs: 10/21