`(2/1.2 + 2/3.4 + ... + 2/99.100) . (x^2 +x+1945)/2 > 1975 . (1/51 + 1/52 + ... + 1/99 + 1/100)`
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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
30 tháng 7 2016
Xét vế trái biểu thức, ta có:
\(\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\right)\cdot x\)
\(=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\cdot x\)
\(=\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\right]\cdot x\)
\(=\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\right]\cdot x\)
\(=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\cdot x\)
Xét vế phải biểu thức, ta có:
\(\frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{99}+\frac{2012}{100}=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\cdot2012\)
Từ đầu bài và 2 kết luận trên, ta suy ra:
\(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\cdot x=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\cdot2012\)
\(\Rightarrow x=2012\)
NH
0
19 tháng 2 2021
Ta có : B=1-1/2+1/3-...+1/99-1/100= ( 1+1/3+...+1/99) -(1/2+....+1/100)= ( 1+1/2+1/3+....+1/99+1/100)-2.(1/2+...+1/100) =1+1/2+1/3+...+1/100 - ( 1+...+1/50) = (1+1/2+...+1/50) + ( 1/51+1/52+...+1/100) - ( 1+...+1/50)= 1/51 +1/52+...+1/100 (1)
C=1/(1.2) +1/(3.4) +...+1/(99.100) = 1-1/2+ 1/3-1/4+...+1/99-1/100 =...
Biểu thức C phần còn lại làm tương tự giống phần (1) nhé => C= 1/51+1/52+...+1/100 (2)
A=1/51+...+1/100(3)
Từ (1),(2) và (3)=>A=B=C (đpcm) . Chúc cậu học tốt !
\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)
\(=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{99}-\dfrac{1}{100}\right)\)
=\(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(\left(\dfrac{2}{1\cdot2}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{99\cdot100}\right)\cdot\dfrac{x^2+x+1945}{2}>1975\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\)=>\(2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\cdot\dfrac{x^2+x+1945}{2}>1975\left(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\)
=>\(x^2+x+1945>1975\)
=>\(x^2+x-30>0\)
=>(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