Question

\(\dfrac{2}{1²}\) . \(\dfrac{6}{2²}\) . \(\dfrac{12}{3³}\) . \(\dfrac{20}{4²}\) +....+ \(\dfrac{110}{10²}\) . x = -20 

Help me

Answer 1

Sửa đề

\(\dfrac{2}{1^2}\cdot\dfrac{6}{2^2}\cdot\dfrac{12}{3^3}\cdot.......\cdot\dfrac{110}{10^2}\cdot x=-20\)

\(\dfrac{2}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\cdot\cdot\cdot\dfrac{11\cdot10}{10\cdot10}\cdot x=-20\)

\(\dfrac{\left(2\cdot3\cdot4\cdot....\cdot11\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot\dfrac{\left(1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot10\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot x=-20\)

\(11\cdot x=-20\\ x=-\dfrac{20}{11}\)

Answer 2

Ko đề cho thêm \(\dfrac{20}{4²}\) mà