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Bài 1:\(A=1-\frac{1}{2}+1-\frac{1}{6}+.......+1-\frac{1}{9900}\)
\(=1-\frac{1}{1.2}+1-\frac{1}{2.3}+........+1-\frac{1}{99.100}\)
\(=99-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\right)=99-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)
\(=99-\left(1-\frac{1}{100}\right)=99-\frac{99}{100}=\frac{9801}{100}\)
Bài 2:\(A=\frac{1}{299}.\left(\frac{299}{1.300}+\frac{299}{2.301}+.........+\frac{299}{101.400}\right)\)
\(=\frac{1}{299}.\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+.........+\frac{1}{101}-\frac{1}{400}\right)\)
\(=\frac{1}{299}.\left(1+\frac{1}{2}+......+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-.......-\frac{1}{400}\right)\)
\(=\frac{1}{299}.\left[\left(1+\frac{1}{2}+.......+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+......+\frac{1}{400}\right)\right]\)(đpcm)
1/
\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{9900}\right)\)
\(=\left(1+1+...+1\right)\left(50so\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\right)\)
\(=50-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(=50-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=50-\left(1-\frac{1}{100}\right)=49+\frac{1}{100}=\frac{4901}{100}\)
2/
\(=\frac{1}{299}\left(\frac{299}{1.300}+\frac{299}{2.301}+...+\frac{299}{101.400}\right)\)
\(=\frac{1}{299}\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}\right)\)
\(=\frac{1}{299}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]\)
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....+\frac{1}{97.100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{97}-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\cdot\frac{99}{100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{33}{100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow x=\frac{0,33\times100}{0,33}=100\)
Câu a)
\(B=\frac{\left(5-\frac{5}{13}+\frac{5}{19}-\frac{5}{27}\right)\cdot4024\cdot4026}{\left(10-\frac{10}{13}+\frac{10}{19}-\frac{10}{27}\right)\cdot2012\cdot2013}\)
\(\Leftrightarrow B=\frac{\left(5\cdot1-5\cdot\frac{1}{13}+5\cdot\frac{1}{19}-5\cdot\frac{1}{27}\right)\cdot2012\cdot2\cdot2013\cdot2}{\left(10\cdot1-10\cdot\frac{1}{13}+10\cdot\frac{1}{19}-10\cdot\frac{1}{27}\right)\cdot2012\cdot2013}\)
\(\Leftrightarrow B=\frac{5\left(1-\frac{1}{13}+\frac{1}{19}-\frac{1}{27}\right)\cdot2\cdot2}{5\cdot2\left(1-\frac{1}{13}+\frac{1}{19}-\frac{1}{27}\right)}\)
\(\Leftrightarrow B=2\)
A=(1/10-1)(1/11-1)...(1/100-1)
A=(-9/10)(-10/11)...(99/100)
vì có 99 số vậy tích trên là âm
a=-9/100
a= 1 chác chắn 100%