(1/1.3+1/3.5+.....+1/99.101) -X =2
Bài 1: Tính tổng
a, 2\1.3+2\3.5+2\5.7+.......+2\99.101
b, 5\1.3+5\3.5+5\5.7+......+5\99.101
Bài 2: CMR phân số 2n+1\3n+2 là phân số tối giản
Bài 1:
Ta có:
\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
\(=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+...+\left(\frac{1}{99}-\frac{1}{101}\right)\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
b, Đặt \(A=\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)
\(\Rightarrow\frac{2}{5}A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
Từ (a) \(\Rightarrow\frac{2}{5}A=\frac{100}{101}\)
\(\Rightarrow A=\frac{100}{101}:\frac{2}{5}=\frac{100}{101}.\text{5/2}=\frac{250}{101}\)
Bài 2:
Đặt \(\left(2n+1;3n+2\right)=d\left(d\inℕ^∗\right)\)
\(\Rightarrow\hept{\begin{cases}2n+1⋮d\\3n+2⋮d\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}3\left(2n+1\right)⋮d\\2\left(3n+2\right)⋮d\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}6n+3⋮d\\6n+4⋮d\end{cases}}\)
\(\Rightarrow\left(6n+4\right)-\left(6n+3\right)⋮d\)
\(\Rightarrow1⋮d\Rightarrow d\inƯ\left(1\right)\Rightarrow d=1\)
\(\Rightarrow\left(2n+1;3n+2\right)=1\)
\(\Rightarrow\frac{2n+1}{3n+2}\)là phân số tối giản
1. Giải
a, \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=2.\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\right)\)
\(=\frac{2}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
b, \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(=5.\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{5.100}{2.101}=\frac{500}{202}=\frac{250}{101}\)
2. Giải
Gọi ước chung lớn nhất của 2n + 1 và 3n + 2 là d (d thuộc N*)
=> 2n + 1 \(⋮\)d ; 3n + 2 \(⋮\)d
=> 3(2n + 1) \(⋮\)d ; 2(3n + 2) \(⋮\)d
=> 6n + 3 \(⋮\)d , 6n + 4 \(⋮\)d
=> (6n + 4) - (6n + 3) \(⋮\)d
=> 1 \(⋮\)d
=> d = 1
Vậy \(\frac{2n+1}{3n+2}\)là phân số tối giản
Tính tổng
a. 1/1.2 + 1/2.3 + 1/3.4 + ...+ 1/2015.2016
b. 2/1.3 + 2/3.5 + 2/5.7 + ...+ 2/99.101
c. 5/1.3 + 5/3.5 + 5/5.7 + ...+ 5/99.101
d. 1/2 + 1/6 + 1/12 + 1/20 + ...+ 1/9900
bn nào trả lời nhanh mik tích, cảm ơn ạ
còn cần không bạn, mk làm cho
A = 1/1.3 - 1/3.5 - 1/5.7 - ... - 1/99.101
\(=\dfrac{1}{3}-\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{3}-\dfrac{1}{2}\cdot\dfrac{98}{303}=\dfrac{1}{3}-\dfrac{49}{303}=\dfrac{101-49}{303}=\dfrac{52}{303}\)
Bài 3 : (1+1/1.3)(1+1/2.4)(1+1/3.5)...(1+1/99.101)
bài 4:
(x-3)(x+5)/(x-2)^2 <0
\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right).....\left(1+\frac{1}{99.101}\right)\)
\(=\frac{2.2}{1.3}\frac{3.3}{2.4}.....\frac{100.100}{99.101}\)
\(=\frac{\left(2.3.4.....100\right).\left(2.3.4.....100\right)}{\left(1.2.3.....99\right).\left(3.4.5.....101\right)}\)
\(=\frac{100.2}{101}=\frac{200}{101}\)
\(\frac{\left(x-3\right)\left(x+5\right)}{\left(x-2\right)^2}< 0\)
\(\Rightarrow\frac{\left(x-3\right)\left(x+5\right)}{\left(x-2\right).\left(x-2\right)}< 0\)
=> ( x - 3 ) . ( x - 5 ) và ( x - 2 ) . ( x - 2 ) trái dấu
Mà ( x - 2 )2 = ( x - 2 ) . ( x - 2 ) ≥ 0 ∀ x
\(\Rightarrow\hept{\begin{cases}\left(x−3\right).\left(x+5\right)< 0\\\left(x−2\right).\left(x−2\right)>0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x< −5;−5< x< 3\\x>2\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x< −5\\2< x< 3\end{cases}}\)
1/1.3+1/3.5+1/5.7+.........+1/99.101
Đặt A=1/1*3+1/3*5+..+1/99*101
A=2/2*(1/1*3+1/3*5+...+1/99*101)
A=1/2*(2/1*3+2/3*5+..+2/99*101)
A=1/2*(1/1-1/3+1/3-1/5+...+1/99-1/100)
A=1/2*(1/1-1/100)
A=1/2*99/100
A=99/200
50/101 nha
Ai chưa có người yêu thì k và kết bạn với mình nhé
\(\frac{1}{1\cdot3}\)+ ... +\(\frac{1}{99\cdot101}\)
2 lần cái này bằng \(\frac{2}{1\cdot3}\)+\(\frac{2}{99\cdot101}\)
= 1/1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/100
=1-1/100
=> cái này bằng 1-1/100 chia 2 = 99/200
Nên nhớ, tao đang học lớp 6 đấy nhé.
1/1.3+1/3.5+1/5.7+...+1/99.101
A=1/1x3+1/3x5+1/5x7+...+1/99x101
gấp cả 2 vế lên 2 lần ta có:
Ax2=2/1x3+2/3x5+2/5x7+...+2/99x101
Ax2=1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101
Ax2=1-1/101
Ax2=100/101
A=100/101:2=50/101
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
Chúc bạn học tốt nha !!!
a)1+1/1.3+1/3.5+...+1/99.101
b)1/2+1/2.5+1/5.8+1/8.11+...+1/98.101
a) đặt
\(S=1+\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{99\cdot101}\\ 2S=2+\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\\ 2S=2+\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ 2S=2+1-\dfrac{1}{101}\\ 2S=\dfrac{302}{101}\\ S=\dfrac{151}{101}\)
b)
đặt
\(S=\dfrac{1}{2}+\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{98\cdot101}\\ 3S=\dfrac{3}{2}+\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{98\cdot101}\\ 3S=\dfrac{3}{2}+\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{98}-\dfrac{1}{101}\\ 3S=\dfrac{3}{2}+\dfrac{1}{2}-\dfrac{1}{101}\\ 3S=\dfrac{201}{101}\\ S=\dfrac{67}{101}\)
\(2A-1=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(2A-1=1-\dfrac{1}{101}=\dfrac{100}{101}\)
\(2A=\dfrac{201}{101}\Rightarrow A=\dfrac{201}{202}\)
Tính 1/1.3+1/3.5+1/5.7+...1/99.101
1/1.3 + 1/3.5 + 1/5.7 + ... + 1/99.101
= 1/2.(2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101)
= 1/2.(1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/99 - 1/101)
= 1/2.(1 - 1/101)
= 1/2.100/101
= 50/101
\(\text{Đặt : }A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{99.101}\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)
\(\Rightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(\Rightarrow2A=1-\frac{1}{101}\)
\(\Rightarrow A=\frac{100}{101}:2=\frac{50}{101}\)
A=1/1.3+1/3.5+1/5.7+....+1/99.101
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{1}-\frac{1}{101}=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)
\(\Rightarrow A=\frac{100}{101}:2=\frac{100}{101}\times\frac{1}{2}=\frac{50}{101}\)