tìm x biêt 1/1.2+1/1.3+...+1/99.100.2014x=1/51+1/52+....+1/100.2013
giúp mk vs
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
L
1
6 tháng 1
\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)
\(=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{99}-\dfrac{1}{100}\right)\)
=\(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(\left(\dfrac{2}{1\cdot2}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{99\cdot100}\right)\cdot\dfrac{x^2+x+1945}{2}>1975\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\)=>\(2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\cdot\dfrac{x^2+x+1945}{2}>1975\left(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\)
=>\(x^2+x+1945>1975\)
=>\(x^2+x-30>0\)
=>(x+6)(x-5)>0
TH1: \(\left\{{}\begin{matrix}x+6>0\\x-5>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>-6\\x>5\end{matrix}\right.\)
=>x>5
TH2: \(\left\{{}\begin{matrix}x+6< 0\\x-5< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< -6\\x< 5\end{matrix}\right.\)
=>x<-6
K
1
11 tháng 12 2021
\(\Leftrightarrow2x\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
Đặt \(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100};B=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
\(\Leftrightarrow A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ \Leftrightarrow A=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\\ \Leftrightarrow A=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\\ \Leftrightarrow A=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}-1-\dfrac{1}{2}-\dfrac{1}{3}-...-\dfrac{1}{50}\\ \Leftrightarrow A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=B\)
\(\Leftrightarrow2x.A=B\Leftrightarrow2x.B-B=0\\ \Leftrightarrow B\left(2x-1\right)=0\\ \Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)
13 tháng 8 2016
Ta có:
\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\)
=> \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right).x=\frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{99}+\frac{2012}{100}\)
=> \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right).x=2012.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\)
=> x = 2012
TN
2
YN
26 tháng 12 2021
Answer:
Mình làm thành tính tỉ số luôn nhé!
\(A=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}}\)
Ta xét \(B=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{2-1}{1.2}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+...+\frac{1}{100}-1-\frac{1}{2}-...-\frac{1}{50}\)
\(=\left(1-1\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+...+\left(\frac{1}{50}-\frac{1}{50}\right)+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}\)
\(\Rightarrow\frac{A}{B}=1\)
TN
8
19 tháng 1 2016
Xét mẫu số: 1/(2x3) + 1/(3x4) + …… + 1/(99x100)
= 1/1 – 1/2 + 1/3 – 1/4 + ......... + 1/99 – 1/100
= (1 + 1/3 + ............ + 1/99) – (1/2 + 1/4 + .......... + 1/100)
= (1 + 1/3 + ............ + 1/99)+(1/2+1/4+1/6+….+1/100) – (1/2+1/4+1/6+ .......... + 1/100)x2
= (1 + 1/2 + 1/3 + 1/4 + ..... + 1/99 + 1/100) – (1 + 1/2 + 1/3 + ....... +1/50 )
= 1/51 + 1/52 + 1/53 + ............. + 1/100 (Đơn giản số trừ)
Vậy: (1/51 + 1/52 + 1/53 + ............. + 1/100) / (1/1x2 + 1/3x4 + .......... + 1/99x100) =
(1/51 + 1/52 + 1/53 + ............. + 1/100) / (1/51 + 1/52 + 1/53 + ............. + 1/100) = 1
NN
0
ND
1
HP
10 tháng 5 2016
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{99.100}\)
\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
\(B=\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{100}\right)\)
\(B=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(B=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{25}\right)\)
\(B=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+....+\frac{1}{100}=A\)
=>A/B=1
LT
0
