Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
cc,you nói thế thì t cx chẳng muốn làm
để ý thấy: từ: \(100\rightarrow-100\) sẽ có 1 thừa số 0
and........
Ok?
làm linh tinh
b\()\)
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 3/4
Tương tự như vậy với câu a\()\)
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 1/2
Ta có :
\(D=\dfrac{100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+.......+\dfrac{99}{100}}\)
\(\Leftrightarrow D=\dfrac{100-1-\dfrac{1}{2}-\dfrac{1}{3}-......-\dfrac{1}{100}}{\dfrac{1}{2}+\dfrac{2}{3}+.....+\dfrac{99}{100}}\)
\(\Leftrightarrow D=\dfrac{99-\dfrac{1}{2}-\dfrac{1}{3}-......-\dfrac{1}{100}}{\dfrac{1}{2}+\dfrac{2}{3}+....+\dfrac{99}{100}}\)
\(\Leftrightarrow D=\dfrac{\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+.....+\left(1-\dfrac{1}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+.......+\dfrac{99}{100}}\)
\(\Leftrightarrow D=\dfrac{\dfrac{1}{2}+\dfrac{2}{3}+........+\dfrac{99}{100}}{\dfrac{1}{2}+\dfrac{2}{3}+......+\dfrac{99}{100}}=1\)
Ta có:
\(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{99}{100}\)
\(\Rightarrow100-1-\dfrac{1}{2}-...-\dfrac{1}{100}=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{99}{100}\)
\(\Rightarrow100=1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3}+...+\dfrac{1}{100}+\dfrac{99}{100}\)
\(\Rightarrow100=1+1+1+...+1\) (\(100\) số \(1\))
\(\Rightarrow100=100\)
Vậy \(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{99}{100}\) (Đpcm)
Ta có :
\(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...............+\dfrac{1}{100}\right)\)
\(=100-1-\dfrac{1}{2}-\dfrac{1}{3}-..................-\dfrac{1}{100}\)
\(=99-\dfrac{1}{2}-\dfrac{1}{3}-................-\dfrac{1}{100}\)
\(=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+..................+\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}+\dfrac{2}{3}+.................+\dfrac{99}{100}\)
Vậy :\(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...............+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+....................+\dfrac{99}{100}\)
\(\Rightarrowđpcm\)
Ta có:\(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}\)
\(=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+...+\left(1-\dfrac{1}{100}\right)\)
\(=\left(1-1\right)+\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+...+\left(1-\dfrac{1}{100}\right)\)\(=\left(1+1+...+1\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)\)
\(=100-\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)\)(đpcm)
dễ
TRẦN MINH HOÀNG