\(=\dfrac{1\cdot2\cdot3+8\cdot1\cdot2\cdot3+64\cdot1\cdot2\cdot3+125\cdot1\cdot2\cdot3}{1\cdot3\cdot4+8\cdot1\cdot3\cdot4+64\cdot1\cdot3\cdot4+125\cdot1\cdot3\cdot4}\)

\(=\dfrac{1\cdot2\cdot3}{1\cdot3\cdot4}=\dfrac{2}{4}=\dfrac{1}{2}\)

x là nhân

Trả lời:

\(\frac{2\times4+2\times4\times8+4\times8\times16+8\times16\times32}{3\times4+2\times6\times8+4\times12\times16+8\times24\times32}\)

\(=\frac{2\times4+2\times4\times8+2.2\times2.4\times16+4.2\times4.4\times32}{3\times4+2\times2.3\times2.4+4\times3.4\times4.4+8\times8.3\times8.4}\)

\(=\frac{2\times4\times\left(1+8+2\times2\times16+4\times4\times32\right)}{3\times4\times\left(1+2\times2\times2+4\times4\times4+8\times8\times8\right)}\)

\(=\frac{2\times4\times\left(1+8+64+512\right)}{3\times4\times\left(1+8+64+512\right)}\)

\(=\frac{2}{3}\)

Đặt \(A=1\cdot2\cdot4+2\cdot3\cdot5+3\cdot4\cdot6+\cdots+100\cdot101\cdot103\)

\(=1\cdot2\cdot\left(3+1\right)+2\cdot3\cdot\left(4+1\right)+\cdots+100\cdot101\cdot\left(102+1\right)\)

\(=\left(1\cdot2\cdot3+2\cdot3\cdot4+\cdots+100\cdot101\cdot102\right)+\left(1\cdot2+2\cdot3+\cdots+100\cdot101\right)\)

Đặt \(B=1\cdot2\cdot3+2\cdot3\cdot4+\cdots+100\cdot101\cdot102\)

\(=\left(2-1\right)\cdot2\cdot\left(2+1\right)+\left(3-1\right)\cdot3\cdot\left(3+1\right)+\cdots+\left(101-1\right)\cdot101\cdot\left(101+1\right)\)

\(=2\left(2^2-1\right)+3\left(3^2-1\right)+\cdots+101\left(101^2-1\right)\)

\(=\left(2^3+3^3+\cdots+101^3\right)-\left(2+3+\cdots+101\right)\)

\(=\left(1^3+2^3+3^3+\cdots+101^3\right)-1-\left(2+3+\cdots+101\right)\)

\(=\left(1^3+2^3+\cdots+101^3\right)-\left(1+2+3+\cdots+101\right)\)

\(=\left(1+2+3+\cdots+101\right)^2-\left(1+2+3+\cdots+101\right)\)

\(=\left\lbrack101\cdot\frac{102}{2}\right\rbrack^2-101\cdot\frac{102}{2}=\left(101\cdot51\right)^2-101\cdot51\)

Đặt \(C=1\cdot2+2\cdot3+\cdots+100\cdot101\)

\(=1\left(1+1\right)+2\left(2+1\right)+\cdots+100\left(100+1\right)\)

\(=\left(1^2+2^2+\cdots+100^2\right)+\left(1+2+\cdots+100\right)\)

\(=\frac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}+\frac{100\cdot101}{2}=\frac{100\cdot101\cdot201}{6}+50\cdot101\)

\(=50\cdot101\cdot67+50\cdot101=50\cdot101\cdot68\)

Ta có: A\(=\left(1\cdot2\cdot3+2\cdot3\cdot4+\cdots+100\cdot101\cdot102\right)+\left(1\cdot2+2\cdot3+\cdots+100\cdot101\right)\)

=B+C

\(=\left(101\cdot51\right)^2-101\cdot51+50\cdot101\cdot68\)

\(=101^2\cdot51^2-101\cdot51+50\cdot101\cdot68=101\left(101\cdot51^2-51+50\cdot68\right)=101\cdot266050\)

Đặt \(D=1\cdot2^2+2\cdot3^2+\cdots+100\cdot101^2\)

\(=2^2\left(2-1\right)+3^2\left(3-1\right)+\cdots+101^2\left(101-1\right)\)

\(=\left(2^3+3^3+\cdots+101^3\right)-\left(2^2+3^2+\cdots+101^2\right)\)

\(=\left(1^3+2^3+\cdots+101^3\right)-\left(1^2+2^2+\cdots+101^2\right)\)
\(=\left(1+2+\cdots+101\right)^2-101\cdot\frac{\left(101+1\right)\left(2\cdot101+1\right)}{6}\)

\(=\left(101\cdot\frac{102}{2}\right)^2-101\cdot17\cdot2023=101^2\cdot51^2-101\cdot17\cdot2023\)

\(=101\cdot17\left(101\cdot17\cdot3^2-2023\right)=101\cdot17\cdot13430\)

Ta có: \(\frac{1\cdot2\cdot4+2\cdot3\cdot5+3\cdot4\cdot6+\cdots+100\cdot101\cdot103}{1\cdot2^2+2\cdot3^2+\cdots+100\cdot101^2}\)

\(=\frac{101\cdot266050}{101\cdot17\cdot13430}=\frac{1565}{1343}\)