Tình : A = (200-2-1)(199-2-1).......(101-2-1)
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XO
7 tháng 8 2020
A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Lại có B = \(\frac{1}{101.200}+\frac{1}{102.199}+...+\frac{1}{200.101}\)
=> 301B = \(\frac{301}{101.200}+\frac{301}{102.199}+...+\frac{301}{200.101}\)
=> 301B = \(\frac{1}{101}+\frac{1}{200}+\frac{1}{102}+\frac{1}{199}+...+\frac{1}{200}+\frac{1}{101}=2\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)\)
=> B = \(\frac{2}{301}\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)\)
Khi đó \(\frac{A}{B}=\frac{\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)}{\frac{2}{301}\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)}=\frac{1}{\frac{2}{301}}=\frac{301}{2}=150,5\)
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Q
18 tháng 7 2017
A=(200-2-1)(199-2-1)....(101-2-1)
\(A=\left(\frac{1}{200^2}-1\right)\left(\frac{1}{199^2}-1\right)\left(\frac{1}{198^2}-1\right).....\left(\frac{1}{101^2}-1\right)\)
\(A=\frac{1-200^2}{200^2}.\frac{1-199^2}{199^2}.\frac{1-198^2}{198^2}.....\frac{1-101^2}{101^2}\)
\(A=\frac{\left(1-200\right)\left(1+200\right)}{200^2}.\frac{\left(1-199\right)\left(1+199\right)}{199^2}.....\frac{\left(1-100\right)\left(1+100\right)}{100^2}.\frac{\left(1-101\right)\left(1+101\right)}{101^2}\)
\(A=\frac{-199.201}{200.200}.\frac{-198.200}{199.199}.\frac{-197.199}{198.198}.....\frac{-99.101}{100.100}.\frac{-100.102}{101.101}\)
\(A=\frac{199.201}{200.200}.\frac{198.200}{199.199}.\frac{197.199}{198.198}.....\frac{99.101}{100.100}.\frac{100.102}{101.101}\)
\(\Rightarrow A=\frac{200}{2.101}=\frac{201}{202}\)
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