Question

Answer 1

giúp mk bài 3 với

 

Answer 2

Bài 3: 

a) Ta có: \(2+4+6+8+...+2x=110\)

\(\Leftrightarrow1+2+3+4+...+x=\dfrac{110}{2}\)

\(\Leftrightarrow x^2+x-110=0\)

\(\Leftrightarrow x^2+11x-10x-110=0\)

\(\Leftrightarrow x\left(x+11\right)-10\left(x+11\right)=0\)

\(\Leftrightarrow\left(x+11\right)\left(x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+11=0\\x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-11\left(loại\right)\\x=10\left(nhận\right)\end{matrix}\right.\)

Vậy: x=10