(1/2+1).(1/3+1).(1/4+1)...(1/999+1)
A=1/1×2+1/3×4+1/4×5+...1/999×1000
B=1/501×1000+1/502×999+...+1/999×502+1/1000×501
Tính A/B
1. Chứng tỏ rằng:
a. 1/n + 1/n+1 = 1/n + 1/n+1
b. 1/1 . 1/2 +1/2 . 1/3+ 1/3 . 1/4+.......+ 1/998 . 1/999+ 1/999. 1/1000
a, Điều đương nhiên
b,\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{999.1000}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.........+\frac{1}{999}-\frac{1}{1000}\)
= \(1-\frac{1}{1000}\)
= \(\frac{999}{1000}\)
1/1*2+1/2*3+1/3*4+.......+1/999*1000+1
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{999.1000}+1\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{999}-\dfrac{1}{1000}+1\)
\(=1-\dfrac{1}{1000}+1\)
\(=\dfrac{1999}{1000}\).
1*1/2*1*1/3*1*1/4*...*1*1/999
Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn.
1/(1×2) + 1/ ( 2×3) +1/ (3×4) +................+1/(999×1000) +1
1/1*2+1/2*3+1/3*4+...+1/999*1000+1=?
1/1*2+1/2*3+,,,,,+1/999*1000+1
=1/1-1/2+1/2-1/3+,,,,+1/999-1/1000+1
=1-1/1000+1
=1+1-1/1000
=2-1/1000
=1999/1000
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}+1\)
Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}\)
\(\Rightarrow A=1-\frac{1}{1000}=\frac{999}{1000}\)
Thay vào ta có : \(\frac{999}{1000}+1=\frac{1999}{1000}\)
Vậy ...
1 / 1x2 + 1 / 2x3 + 1/ 3x4 +....+ 1/999x1000 + 1
= ( 1 / 1 x2 + 1/ 2 x3 + 1/ 3x4 + ....+ 1/999x 1000 ) + 1
= ( 1 x1 / 1 x2 + 1x2 / 2x3 + 1x 3 / 3x4 + ..... 1x 999 / 999x 1000 ) + 1
= ( 1 - 1 / 2 + 1 / 2 - 1 / 3 - 1 / 3 - 1 / 4 + ........+ 1 / 999 - 1 / 1000 ) + 1
= ( 1 - 1 / 1000 ) + 1
= 999 / 1000 + 1
= 1999 / 1000
Tk tớ nha
1/1*2+1/2*3+1/3*4+.....+1/999*1000+1
ta thấy 1/1-1/2=1/1x2
vậy tích đó \(1-\frac{1}{2}+\frac{1}{2}.............-\frac{1}{1000}=1-\frac{1}{1000}=\frac{999}{1000}\)
1/1*2 +1/2*3 + 1/3*4 +..........+1/999*1000 +1
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{999\cdot1000}+1\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}+1\)
\(=1-\frac{1}{1000}+1\)
\(=\frac{999}{1000}+1=\frac{1999}{1000}\)
1/1*2 + 1/2*3 + 1/3*4 +................+1/999*1000 +1
=1/2 + 1/6 + 1/12 + ....................+1/999000 +1
=1-1/2 + 1/2-1/3 + 1/3-1/4 +...+ 1/999- 1/1000+1
=1- 1/1000+1
=999/1000+1
=1999/1000
1/1*2+1/2*3+1/3*4+..........+1/999*1000+1
1/1*2+1/2*3+1/3*4+1/4*5+...+1/999*1000+1
=1-1/1000+1
=1000+1-1/1000+1
=1000/1001
1/1*2+1/2*3+1/3*4+...+1/999*1000+1