Bài 1: Tính nhanh
5/3+5/3x5+5/5x7+...+5/101x103
Tính : 5/3 + 5/3x5 + 5/5x7 + ... + 5/101x103 ai xong mik tick
Tính bằng cách thuận tiện nhất :
a) 9+ 9/2 + 9/4 + 9/8 + ... +9/128+ 9/256
b) 5/3 + 5/3x5 + 5/5x7 + ... + 5/101x103
b ) Đặt \(A=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{101.103}\)
\(\Rightarrow A=\frac{5}{2}\left(\frac{5}{1}-\frac{5}{3}+\frac{5}{3}-\frac{5}{5}+....+\frac{5}{101}-\frac{5}{103}\right)\)
\(\Rightarrow A=\frac{5}{2}\left(5-\frac{5}{103}\right)\)
trình bày bài giải tính nhanh
1/1x3+1/3x5+1/5x7+1/7x9...........1/101x103
\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.+\frac{1}{101}-\frac{1}{103}\right)\)
\(\frac{1}{2}\left(1-\frac{1}{103}\right)=\frac{1}{2}\cdot\frac{100}{103}=\frac{50}{103}\)
xong r đó
Ta có:
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{101.103}\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{101.103}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{103}\right)=\frac{50}{103}\)
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{101\cdot103}\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{103}\right)\)
\(=\frac{1}{2}\cdot\frac{102}{103}=\frac{102}{206}\)
\(=\frac{51}{103}\)
các bạn ơi giúp mình với !!!!!
Bài 1: 1+ 1/5 + 2/15 + 1/125 + 1/625 + 1/3125
Bài 2: 3/1x3 + 3/3x5 + 3/5x7 + .... + 3/99x101 + 3/101x103
MÌNH CẦN GẤP . CÁC BẠN GỬI CHO MÌNH TRƯỚC 1 GIỜ NHÉ
CẢM ƠN
2) \(\frac{3}{1\times3}+\frac{3}{3\times5}+\frac{3}{5\times7}+...+\frac{3}{99\times101}+\frac{3}{101\times103}\)
\(=\frac{3}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{101\times103}\right)\)
\(=\frac{3}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=\frac{3}{2}\times\left(1-\frac{1}{103}\right)\)
\(=\frac{3}{2}\times\frac{101}{103}\)
\(=\frac{303}{206}\)
Tính
a. 3/(3x5) + 3/(5x7) + 3/(7x9) +... + 3/(99x101)
b. 5/(3x5) +5/(5x7) +5/(7x9) +...+ 5/(99x101)
917749738461936926399639748776398646491639394748947630373937366
Tính tổng:
A=1/1+1/2+1/3+1/4+1/5+1/6
B=1x3+3x5+5x7+7x9+...+95x97+97x99
(1/1x3+1/3x5+1/5x7+1/7x9+1/9x11+1/11x13)xy=3/5
\(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+\frac{1}{11\times13}\)
\(=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}+\frac{2}{11\times13}\right)\)
\(=\frac{1}{2}\times\left(\frac{3-1}{1\times3}+\frac{5-3}{3\times5}+\frac{7-5}{5\times7}+\frac{9-7}{7\times9}+\frac{11-9}{9\times11}+\frac{13-11}{11\times13}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{13}\right)=\frac{6}{13}\)
Do đó ta có:
\(\frac{6}{13}\times y=\frac{3}{5}\)
\(\Leftrightarrow y=\frac{13}{10}\).
ngu các em học quá ngu
help me !!!
bài 15
a) 2/1x3 + 2/3x5 + 2/5.7+......+2/99x101
b) 5/1x3 + 5/3x5 + 5/5x7+......+5/99x101
bài 16
chứng tỏ rằng phân số 2n+1/3n+1 là phân số tối giản
bạn nào làm đc đầu tiên mk tick nha
a)\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
= \(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
= \(1-\dfrac{1}{101}\)
=\(\dfrac{100}{101}\)
\(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+...+\dfrac{5}{99.101}\)
=\(\dfrac{5}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99+101}\right)\)
=\(\dfrac{5}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
=\(\dfrac{5}{2}.\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{5}{2}-\dfrac{100}{101}\)
= \(\dfrac{305}{202}\)
Bài 16:
A = \(\dfrac{2n+1}{3n+1}\); đkxđ n \(\ne\) - \(\dfrac{1}{3}\)
Gọi ước chung lớn nhất của 2n + 1 là d
Ta có: 2n + 1 ⋮ d; 3n + 1 ⋮ d
2n + 1 ⋮ d ⇒ 3.(2n + 1) ⋮ d ⇒ 6n + 3 ⋮ d
3n + 1 ⋮ d ⇒ 2.( 3n+ 1) ⋮ d ⇒ 6n + 2 ⋮ d
⇒ 6n + 3 - (6n + 2) ⋮ d ⇒ 6n + 3 - 6n - 2⋮ d
⇒ 1 ⋮ d ⇒ d = 1
Ước chung lớn nhất của 2n + 1 và 3n + 1 là 1
Hay 2n + 1 và 3n + 1 là hai số nguyên tố cùng nhau (đpcm)
5/1x3+5/3x5+5/5x7+...+5/47x49