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Sửa đề: \(A=\frac{1\cdot98+2\cdot97+\cdots+98\cdot1}{1\cdot2+2\cdot3+\cdots+98\cdot99}\)

Đặt \(B=1\cdot98+2\cdot97+\cdots+98\cdot1\)

\(=2\left(1\cdot98+2\cdot97+\cdots+49\cdot50\right)\)

\(=2\cdot\left\lbrack1\left(99-1\right)+2\left(99-2\right)+\cdots+49\left(99-49\right)\right\rbrack\)

\(=2\cdot\left\lbrack99\left(1+2+\cdots+49\right)-\left(1^2+2^2+\cdots+49^2\right)\right\rbrack\)

\(=2\cdot\left\lbrack99\cdot\frac{49\cdot50}{2}-\frac{49\left(49+1\right)\left(2\cdot49+1\right)}{6}\right\rbrack=2\cdot\left\lbrack99\cdot49\cdot25-49\cdot25\cdot33\right\rbrack\)

\(=2\cdot25\cdot49\cdot33\left(3-1\right)=50\cdot49\cdot33\cdot2=100\cdot49\cdot33\)

Đặt \(C=1\cdot2+2\cdot3+\cdots+98\cdot99\)

\(=1\left(1+1\right)+2\left(2+1\right)+\cdots+98\left(98+1\right)\)

\(=\left(1^2+2^2+\cdots+98^2\right)+\left(1+2+\cdots+98\right)\)

\(=\frac{98\left(98+1\right)\left(2\cdot98+1\right)}{6}+\frac{98\cdot99}{2}=49\cdot33\cdot197+49\cdot99=49\cdot33\cdot\left(197+3\right)=49\cdot33\cdot200\)

Ta có: \(A=\frac{1\cdot98+2\cdot97+\cdots+98\cdot1}{1\cdot2+2\cdot3+\cdots+98\cdot99}\)

\(=\frac{100\cdot49\cdot33}{49\cdot33\cdot200}=\frac{100}{200}=\frac12\)

Tìm x, biết: \(\left ( \frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110} \right ).x = \frac{1}{1.11}+\frac{1}{1.12}+...+\frac{1}{100.110}\)- Trường Toán Trực tuyến Pitago – Giải pháp giúp em học toán vững vàng!

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