Chứng tỏ rằng
[200-(3+2/3+2/4+2/5+...+2/100]:[1/2+2/3+3/4+...+99/100]=2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
26 tháng 4 2017
Ta có :
\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{2}{5}+............+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+.................+\dfrac{99}{100}}\)
\(=\dfrac{200-2-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+.............+\dfrac{2}{100}\right)}{1-\dfrac{1}{2}+1-\dfrac{1}{3}+............+1-\dfrac{1}{100}}\)
\(=\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+...........+\dfrac{2}{100}\right)}{\left(1+1+.........+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+........+\dfrac{1}{100}\right)}\)
\(=\dfrac{2.\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+..........+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+.........+\dfrac{1}{100}\right)}\)
\(=2\)
Vậy \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+..........+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+........+\dfrac{99}{100}}=2\rightarrowđpcm\)
6 tháng 5 2017
Ta có \(A=\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+\frac{2}{5}+....+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+......+\frac{99}{100}}\)
\(A=\frac{200-2\left(\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{100}\right)}{\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{100}\right)}\)
\(A=\frac{2\left[100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+.....+\frac{1}{100}\right)\right]}{100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{100}\right)}\)
\(\Rightarrow A=2\)
NM
1
LN
1
6 tháng 5 2017
Ta có
200-(3+2/3+...+2/100)
=200-(3+2(1/3+...+1/100)
=200-(3+2 (1-2/3+1-3/4+...+1-99/100))
=200-(3+2(98-(2/3+3/4+...+99/100)))
=200-3-196-(2/3+3/4+...+99/100)
=1-(2/3+3/4+...+99/100)
Thay:1-(2/3+3/4+...+99/100)/2/3+3/4+......+99/100=1/(1/2)=2
* Bỏ ngoặc vuông đi :(
\(\text{Ta có:}\)
\(200-\left(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\right)\)
\(\rightarrow200-2-\left(1+\frac{2}{3}+...+\frac{2}{100}\right)\)
\(\rightarrow198-\left(1+\frac{2}{3}+...+\frac{2}{100}\right)\)
\(\rightarrow198-\left(1+\frac{2}{3}+...+\frac{2}{100}\right)\)
\(\rightarrow2.[99-\left(\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}\right)]\) \(\left(1\right)\)
\(\text{Ta có:}\)
\(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)
\(\text{Rút}\)\(\left(1\right)\)\(\text{ra có 99 số}\)
\(\rightarrow99-\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\) \(\left(2\right)\)
\(\text{Từ}\)\(\left(1\right)\)\(\text{và}\)\(\left(2\right)\)\(\Rightarrow\)\(200-\left(3+\frac{2}{3}+\frac{2}{4}+\frac{2}{5}+...+\frac{2}{100}\right):\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\right)=2\)