tính D =\(\dfrac{100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+....+\dfrac{99}{100}}\)
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28 tháng 3 2017
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)
AH
Akai Haruma
Giáo viên
5 tháng 7 2021
Lời giải:
Gọi phân số vế trái là $A$. Gọi tử số là $T$. Xét mẫu số:
\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(=1-\frac{1}{2}+1-\frac{1}{3}+1-\frac{1}{4}+....+1-\frac{1}{100}\)
\(=99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=100-(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100})\)
\(=\frac{1}{2}\left[200-(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100})\right]=\frac{1}{2}T\)
$\Rightarrow A=\frac{T}{\frac{1}{2}T}=2$
Ta có đpcm.
6 tháng 7 2021
Giải:
Vì \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}=2\) nên phần tử gấp 2 lần phần mẫu
Ta có:
\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)
\(=\dfrac{2.\left[100-\left(\dfrac{3}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{100}\right)\right]}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)
\(=\dfrac{2.\left[\left(2-\dfrac{3}{2}\right)+\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{5}\right)+...+\left(1-\dfrac{1}{100}\right)\right]}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)
\(=\dfrac{2.\left(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{4}{5}+...+\dfrac{99}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)
\(=2\)
Vậy \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}=2\left(đpcm\right)\)
Chúc bạn học tốt!
16 tháng 3 2017
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)
27 tháng 6 2022
1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)
2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)
Ta có :
\(D=\dfrac{100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+.......+\dfrac{99}{100}}\)
\(\Leftrightarrow D=\dfrac{100-1-\dfrac{1}{2}-\dfrac{1}{3}-......-\dfrac{1}{100}}{\dfrac{1}{2}+\dfrac{2}{3}+.....+\dfrac{99}{100}}\)
\(\Leftrightarrow D=\dfrac{99-\dfrac{1}{2}-\dfrac{1}{3}-......-\dfrac{1}{100}}{\dfrac{1}{2}+\dfrac{2}{3}+....+\dfrac{99}{100}}\)
\(\Leftrightarrow D=\dfrac{\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+.....+\left(1-\dfrac{1}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+.......+\dfrac{99}{100}}\)
\(\Leftrightarrow D=\dfrac{\dfrac{1}{2}+\dfrac{2}{3}+........+\dfrac{99}{100}}{\dfrac{1}{2}+\dfrac{2}{3}+......+\dfrac{99}{100}}=1\)
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