tính A=1/3.5+1/5.7 +1/7.9 + ... +1/37.39
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
Tính các tổng sau bằng phương pháp hợp lí nhất:
A= 1/1.2 + 1/2.3 + 1/3.4+...+ 1/49.50
B= 2/3.5 + 2/5.7 + 2/7.9+...+ 2/37.39
A=1/1-1/2+1/2-1/3+1/3-1/4+....+1/49-1/50
A=1/1-1/50
A=49/50
Vay A=49/50
B=1/3-1/5+1/5-1/7....+1/37-1/39
B=1/3-1/39
b=36/117
B=4/13
bài 1 tính
a, A= 2/3.5 + 2/5.7 + 2/7.9 + ......+ 2/37.39
b, B= 3/4.7 + 3/7.10 + 3/10.13 +.......+3/73.76
a) \(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)
\(A=\frac{1}{3}-\frac{1}{39}\)
\(A=\frac{13}{39}-\frac{1}{39}=\frac{12}{39}\)
b) \(B=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)
\(B=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\)
\(B=\frac{1}{4}-\frac{1}{76}\)
\(B=\frac{19}{76}-\frac{1}{76}=\frac{18}{76}=\frac{9}{38}\)
a, \(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)
\(A=\frac{1}{3}-\frac{1}{39}\)
\(A=\frac{4}{13}\)
b, \(B=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)
\(B=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\)
\(B=\frac{1}{4}-\frac{1}{76}\)
\(B=\frac{9}{38}\)
tính B = 2/3.5 + 2/5.7 + 2/7.9 + ... + 2/37.39
\(B=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{37.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{37}-\frac{1}{39}=\frac{1}{3}-\frac{1}{39}=\frac{12}{39}=\frac{4}{13}\)
tính: A=1 phần 1.2 +1 phần 2.3+ 1 phần 3.4+...1 phần 49.50
B=2 phần 3.5+ 2 phần 5.7+ 2 phần 7.9 +...+ 2 phần 37.39
C= 3 phần 4.7 + 3 phần 7.1 + 3 phần 10.13 + ... + 3 phần 73.76
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
= \(1-\frac{1}{50}=\frac{49}{50}\)
B = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)
= \(2\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\right)\)
= \(2.\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\right)\)
= \(\frac{2}{2}\left(\frac{1}{3}-\frac{1}{39}\right)\)
= \(\frac{4}{13}\)
C = \(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)
= \(3\left(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\right)\)
= \(3.\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\right)\)
= \(\frac{3}{3}\left(\frac{1}{4}-\frac{1}{76}\right)\)
= \(\frac{9}{38}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
Tính nhanh
1, A = 1/1.2 + 1/2.3 + 1/3.4 + 1/3.4 + ... + 1/49.50
2, B = 2/3.5 + 2/5.7 + 2/7.9 + ... + 2/37.39
3, C = 5^2/1.6 + 5^2/6.11 + ... + 5/26.31
4, D = 1/7 + 1/91 + 1/247 + 1/475 + 1/775 + 1/1147
x + 25 = 64
x = 64 - 25
x = 39
Vậy x = 39
Tính nhanh: C=\(\frac{1}{3.5}\)+\(\frac{1}{5.7}\)+\(\frac{1}{7.9}\)+...+\(\frac{1}{37.39}\)
\(C=\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{37\cdot39}\)
\(2C=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{37\cdot39}\)
\(2C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\)
\(2C=\dfrac{1}{3}-\dfrac{1}{39}\)
\(2C=\dfrac{4}{13}\)
\(C=\dfrac{2}{13}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\)
\(C=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{39}\right)\)
\(C=\frac{1}{2}.\frac{12}{39}\)
\(C=\frac{4}{26}=\frac{2}{13}\)
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 ! >_<