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Ta có 1/1.2-1/2.3=2/1.2.3;1/2.3-1/3.4=2/2.3.4 .....1/98.99-1/99.100=2/98.99.100 2A=2/1.2.3+2/2.3.4+....+2/98.99.100 = 1/1.2-1/2.3+1/2.3-1/3.4+...+1/98.99-1/99.100 = 1/2-1/99.100 = 4949/9900 A =4949/19800
2Q = 1-1/3-1/2+1/4+1/3-1/5-1/4+1/6-........+1/97-1/99-1/98+1/100 = 1-1/2-1/99+1/100 = 4949/9900 >> Q = 49499/19800
\(Q=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+...+\frac{1}{97.99}-\frac{1}{98.100}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}-\frac{1}{2}+\frac{1}{4}+\frac{1}{3}+\frac{1}{5}-\frac{1}{4}+\frac{1}{6}+...+\frac{1}{97}-\frac{1}{99}-\frac{1}{98}+\frac{1}{100}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{100}\right)=\frac{1}{2}.\frac{99}{100}=\frac{99}{200}\) (không chắc cho lắm :v)
Đặt \(A=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{97.99}+\frac{1}{99.100}\)
\(\Rightarrow2A=2\left(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{97.99}+\frac{1}{98.100}\right)\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{2.4}+\frac{2}{3.5}+...+\frac{2}{97.99}+\frac{2}{98.100}\)
\(\Rightarrow2A=\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)+\left(\frac{2}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
\(\Rightarrow2A=\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)+\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(\Rightarrow2A=\left(1-\frac{1}{99}\right)+\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(\Rightarrow2A=\left(\frac{99}{99}-\frac{1}{99}\right)+\left(\frac{50}{100}-\frac{1}{100}\right)\)
\(\Rightarrow2A=\frac{98}{99}+\frac{49}{100}=\frac{9800}{9900}+\frac{4851}{9900}=\frac{14651}{9900}\)
\(\Rightarrow A=\frac{14651}{9900}:2=\frac{14651}{9900}.\frac{1}{2}=\frac{14651}{19800}\)
bạn nhớ thử lại nhé :)
\(a=\frac{6.2.3.4+6.3.4.5+6.4.5.6+...+6.98.99.100}{2.3.4+3.4.5+4.5.6+...+98.99.100}=6\)
thay vào p(x) suy ra a không là nghiệm của đa thức nhé bạn
\(B=\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)\)
\(B=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}...\frac{9898}{99.100}\)
\(B=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{98.101}{99.100}\)
\(B=\frac{1.2.3...98}{2.3.4...99}.\frac{4.5.6...101}{3.4.5...100}\)
\(B=\frac{1}{99}.\frac{101}{3}=\frac{101}{297}\)
