Question

Tim k thuoc N biet, \(x^3y^5+3x^3y^5+.....+\left(2k-1\right)x^3y^5=3249x^3y^5\)

Answer 1

\(x^3y^5+3x^3y^5+...+\left(2k-1\right)x^3y^5=3249x^3y^5\)

\(\Leftrightarrow x^3y^5\left[1+2+3+...+\left(2k-1\right)\right]=3249x^3y^5\)

\(\Leftrightarrow1+3+5+...+\left(2k-1\right)=3249\)

\(\Leftrightarrow\frac{\left[\left(2k-1\right)+1\right].\left(\frac{\left(2k-1\right)-1}{2}+1\right)}{2}=3249\)

\(\Leftrightarrow\frac{2k.\left(k-1+1\right)}{2}=3249\)

\(\Leftrightarrow\frac{2k^2}{2}=3249\)

\(\Leftrightarrow k^2=3249=57^2\) ( ko xét k = - 57 vì theo quy luật thi k luôn dương )

\(\Rightarrow k=57\)

Answer 2

kết quả k = 57