3.3+6.6+9.9+....+90.90
4/3.5+4/5.7+4/7.9+.....4/37.39
B = 2/3.5 + 2/5.7 + 2/7.9 + ... + 2/37.39
Tham khảo :Câu hỏi của hoàng quỳnh dương - Toán lớp 7 - Học toán với OnlineMath
\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)
\(=\frac{1}{3}-\frac{1}{39}\)
\(=\frac{4}{13}\)
Study well ! >_<
b=2/3.5+2/5.7+2/7.9+...+2/37.39
\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)
\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{37}-\frac{1}{39}\)
\(B=\frac{1}{3}-\frac{1}{39}=\frac{13}{39}-\frac{1}{39}=\frac{12}{39}\)
\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)
\(\Rightarrow B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{37}-\frac{1}{39}\)
\(\Rightarrow B=\frac{1}{3}-\frac{1}{39}=\frac{12}{39}\)
tính A=1/3.5+1/5.7 +1/7.9 + ... +1/37.39
✫¸.•°*”˜˜”*°•✫ Ṱђầภ Ḉђết ✫•°*”˜˜”*°•.¸✫ nhân A với 2 rồi phân tích như vậy được
\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+....+\frac{1}{37\cdot39}\)
\(2A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+....+\frac{2}{37\cdot39}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-......+\frac{1}{37}-\frac{1}{39}\)
\(2A=\frac{1}{3}-\frac{1}{39}=\frac{12}{39}=\frac{4}{13}\)
\(A=\frac{4}{13}:2=\frac{4}{13}\cdot\frac{1}{2}=\frac{2}{13}\)
Vậy \(A=\frac{2}{13}\)
Bài làm
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)
\(A=\frac{1}{3}-\frac{1}{39}\)
\(A=\frac{13}{39}-\frac{1}{39}=\frac{12}{39}\)
Vậy \(A=\frac{12}{39}\)
A =1/3.5+1/5.7 +1/7.9 + ... +1/37.39
2.A = 2/3.5+ 2/5.7+ ....+2/37.39
2.A = 1/3- 1/5+1/5-1/7 +....+1/37-1/39
= 1/3- 1/39
2.A= 13/39 -1/39
2.A= 12/39
A = 12/39: 2
A= 2/13
Vậy A= 2/13
tính A=1/3.5+1/5.7 +1/7.9 + ... +1/37.39
=>2A=1/3-1/5+1/5-1/7+1/7-1/9+...+1/37-1/39
=>2A=1/3-1/39=4/13
=>A=2/13
tính B=1/3.5+1/5.7+1/7.9+...+1/37.39
B=1/3.5+1/5.7+1/7.9+...+1/37.39
=1/2(2/3.5+2/5.7+2/7.9+...+2/37.39)
=1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/37-1/39)
=1/2(1/3-1/39)
=1/2(13/39-1/39)
=1/2.4/13
=2/13
1/3.5+1/5.7+1/7.9+....+1/37.39
=1/2.(1/3-1/5+1/5-1/7+1/7-1/9+....+1/37-1/39)
=1/2.(1/3-1/39)
=1/2.4/13
2/13
**** bạn
tìm B biết B=2^2/3.3^3/8^2.4^4/15^3.5^5/24^4.6^6/35^5.7^7/48^6.8^8/63^7.9^9/80^8
tính B = 2/3.5 + 2/5.7 + 2/7.9 + ... + 2/37.39
\(B=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{37.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{37}-\frac{1}{39}=\frac{1}{3}-\frac{1}{39}=\frac{12}{39}=\frac{4}{13}\)
bài 1 tính
A = \(\frac{2}{3.5}+\frac{2}{5.7}+..............+\frac{2}{37.39}\); B = \(\frac{4}{5.7}+\frac{4}{7.9}+..........+\frac{4}{59.61}\) ; C = \(\frac{4}{5.9}+\frac{4}{9.13}+.................+\frac{4}{41.45}\) Bài 2 chứng minh : \(\frac{m}{b.\left(b+m\right)}=\frac{1}{b}-\frac{1}{b+m}\)
4/3.5 + 4/5.7 + 4/7.9 + ... + 4 /97.99 = ?
\(\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+...+\frac{4}{97\cdot99}\)
\(=\frac{2\cdot2}{3\cdot5}+\frac{2\cdot2}{5\cdot7}+\frac{2\cdot2}{7\cdot9}+...+\frac{2\cdot2}{97\cdot99}\)
\(=\frac{2}{3}+\frac{2}{5}-\frac{2}{5}+\frac{2}{7}-\frac{2}{7}+\frac{2}{9}-...+\frac{2}{97}-\frac{2}{99}\)
\(=\frac{2}{3}-\frac{2}{99}\)
\(=\frac{64}{99}\)