Tính : ( 1/ 1 x 2 + 1/ 3 x 4 + ... + 1/ 99 x 100 ) - ( 1/51 + 1/52 + 1/100 )
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VT
1
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
DT
0
NT
1
24 tháng 4 2018
Ta có \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)=\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{50}\right)=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
\(\Rightarrow\text{Đ}PCM\)
NV
0
L
1
11 tháng 4 2023
Sửa đề: \(\dfrac{\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
\(=\dfrac{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
=1
NV
0
Đặt \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{99.100}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(\Rightarrow A=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)
\(\Rightarrow A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(\Rightarrow A-\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)=0\)
\(\Rightarrow\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)-\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)=0\)