A = 1/3 + 1/15 + 1/35 + 1/63 +1/99 + .............. + 1/4088483 là dạng gì lớp mấy
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Đây là dạng tính nhanh tổng các phân số mà mỗi phân số có tử số bằng hiệu hai thừa số dưới mẫu mà thừa số thứ nhất của mẫu này là thừa số thứ hai của mẫu kia em nhé
Có vẻ bạn bị sai đề bài ở chỗ 4088403 nếu là 4088483 sẽ giải được
\(A=\dfrac{1}{3}+\dfrac{1}{3.5}+\dfrac{1}{7.5}+\dfrac{1}{7.9}+...+\dfrac{1}{2021.2023}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}+...+\dfrac{1}{2021}-\dfrac{1}{2023}\)
\(=1-\dfrac{1}{2023}\)
\(=\dfrac{2022}{2023}\)
Giải:
Đặt A = 1/3+1/15+1/35+1/63+1/99+1/143+1/195
2A= 2/(1.3) + 2/(3.5) + 2/(5.7) + 2/(7.9)+2/(9.11) + 2/(11.13)+2/(13.15)
2A=1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9...
2A=1/1-1/15=14/15
Vậy A=14/15 : 2 = 7/15
Nhấn đúng mk nha Tran Dan
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+..+\frac{1}{143}+\frac{1}{195}\)
=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{13.15}\)
= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+..+\frac{1}{13}-\frac{1}{15}\)
= \(1-\frac{1}{15}=\frac{14}{15}\)
tick đúng nha
a) \(A=\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.10}+\dfrac{1}{143}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)+\dfrac{1}{143}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{100}\right)+\dfrac{1}{143}=\dfrac{1}{2}.\dfrac{99}{100}+\dfrac{1}{143}=\dfrac{99}{200}+\dfrac{1}{143}=\dfrac{99.143+200.1}{200.143}=\dfrac{14157+200}{28600}=\dfrac{14357}{28600}\)
b) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=14950\)
\(\Rightarrow x+x+...+x+\left(1+2+...+99\right)=14950\)
\(\Rightarrow100x+\left(\left(99+1\right):2\right).99:2=14950\)
\(\Rightarrow100x+2475=14950\Rightarrow100x=12475\Rightarrow x=\dfrac{12475}{100}=\dfrac{499}{4}\)
1/15 + 1/35 + 1/63 + 1/99 + ... + 1/9999 =
= 1/(3x5) + 1/(5x7) + 1/(7x9) + ... + 1/(99x101)
= (1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ...+ 1/99 - 1/101) : 2
= (1/3 - 1/101) : 2
= 98/303 : 2
= 49/303
tick nha
Đặt phép tính cần tìm là A
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
\(2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)
\(2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)
\(2A=1-\dfrac{1}{13}\)
\(2A=\dfrac{12}{13}\)
\(A=\dfrac{6}{13}\)
\(A=\dfrac{1}{3}+\dfrac{1}{15}+...+\dfrac{1}{143}\\ =\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+...+\dfrac{1}{11\times13}\\ =\dfrac{1}{2}\times\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+...+\dfrac{1}{11\times13}\right)\\ =\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)\\ =\dfrac{1}{2}\times\dfrac{12}{13}\\ =\dfrac{6}{13}\)
P=1/3+1/15+1/35+1/63+1/99
=1:3+1:15+1:35+1:63+1:99
=1:(3+15+35+63+99)
=1:215
=1/215
Vậy:P=1/215
1/3+1/15+1/35+1/63+1/99+……+1/9999
=1/(1×3)+1/(3×5)+1/(5×7)+1/(7×9)+1/(9×11)+……+1/(99×101)
=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)+1/2(1/7-1/9)+1/2(1/9-1/11)+……+1/2(1/99-1/101)
=1/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+……+1/99-1/101)
=1/2(1-1/101)
=1/2×(100/101)
=50/101
1/3+1/15+1/35+1/63+1/99+……+1/9999
=1/(1×3)+1/(3×5)+1/(5×7)+1/(7×9)+1/(9×11)+……+1/(99×101)
=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)+1/2(1/7-1/9)+1/2(1/9-1/11)+……+1/2(1/99-1/101)
=1/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+……+1/99-1/101)
=1/2(1-1/101)
=1/2×(100/101)
=50/101
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{9.11}\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{9.11}\)(tắt 1 bước nha)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{9}-\frac{1}{11}\)
\(2A=\frac{1}{3}-\frac{1}{11}\)
\(2A=\frac{8}{33}\)
\(\Rightarrow A=\frac{4}{33}\)
Vậy A=_____________
LLớp 4
dạng j